Free Math Worksheets

- What does parallelism imply about direction vectors? - What two components are needed to write a parametric equation? - How is the given point used in the line equation?

1. Parallel lines can use the same direction vector, \(\begin{pmatrix}4\\0\\-3\end{pmatrix}\). 2. Use \(P\) as the position vector. Therefore, \(h:\vec{x}=\begin{pmatrix}5\\-2\\0\end{pmatrix}+t\begin{pmatrix}4\\0\\-3\end{pmatrix}\).

\(h:\vec{x}=\begin{pmatrix}5\\-2\\0\end{pmatrix}+t\begin{pmatrix}4\\0\\-3\end{pmatrix}\), where \(t\in\mathbb{R}\)

- How can you write a parametric equation of a line through two given points? - What direction vector can you form from the two points? - How can you test whether a point satisfies a parametric equation? - What must be true about the parameter value in all three coordinate equations?

1. Find a direction vector: \(\vec{v}=\overrightarrow{AB}=\begin{pmatrix}2\\-4\\4\end{pmatrix}\). Therefore, \(g\) can be written as \(\vec{x}=\begin{pmatrix}1\\4\\-2\end{pmatrix}+r\begin{pmatrix}2\\-4\\4\end{pmatrix}\). 2. Test \(C\). From \(1+2r=0\), \(r=-\frac{1}{2}\). Then \(4-4\left(-\frac{1}{2}\right)=6\) and \(-2+4\left(-\frac{1}{2}\right)=-4\). The same parameter value works in all three coordinates, so \(C\) lies on \(g\). 3. Test \(D\). From \(1+2r=4\), \(r=\frac{3}{2}\). This gives \(4-4\left(\frac{3}{2}\right)=-2\), but \(-2+4\left(\frac{3}{2}\right)=4\ne5\). Therefore, \(D\) does not lie on \(g\).

Point \(C\) lies on line \(g\), and point \(D\) does not lie on line \(g\).

- Write a parametric equation for all points on line \(h\). - How can one known coordinate be used to find the parameter value? - What must then be true of the other coordinates? - Is checking only one coordinate enough to confirm that a point lies on a line?

1. A direction vector is \(\overrightarrow{PQ}=\begin{pmatrix}3\\3\\-3\end{pmatrix}\), so \(h\) can be written as \(\vec{x}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+t\begin{pmatrix}3\\3\\-3\end{pmatrix}\). 2. For \(R\), the \(y\)-coordinate gives \(-1+3t=5\), so \(t=2\). The \(z\)-coordinate checks because \(5-3\cdot2=-1\). Thus \(a=2+3\cdot2=8\). 3. For \(S\), the \(x\)-coordinate gives \(2+3t=-1\), so \(t=-1\). Then \(-1+3\cdot(-1)=-4\) and \(5-3\cdot(-1)=8\). The same parameter value works in all coordinates, so \(S\) lies on \(h\).

a) \(a=8\) b) Point \(S\) lies on line \(h\).

- How do the components of the direction vector determine the slope? - What point on the line is given by the position vector? - How do you find where a graph crosses the \(x\)-axis?

1. The slope is the ratio of the changes in \(y\) and \(x\): \(m=\frac{-6}{3}=-2\). 2. The point \((-2, 5)\) lies on the line. Substitute it into \(y=-2x+b\): \(5=-2\cdot(-2)+b\), so \(b=1\). Therefore, \(f(x)=-2x+1\). 3. Set \(f(x)=0\) to find the \(x\)-intercept: \(0=-2x+1\), so \(x=\frac{1}{2}\).

Function: \(f(x)=-2x+1\) \(x\)-intercept: \(\left(\frac{1}{2}, 0\right)\)

- What is the general form of a parametric equation of a line? - What roles do the position vector and direction vector play? - How can you find a direction vector from two points? - Does choosing the other point as the position vector change the line?

1. Find a direction vector for each line by subtracting coordinates: \(\overrightarrow{AB}=\begin{pmatrix}-3\\4\end{pmatrix}\), \(\overrightarrow{BC}=\begin{pmatrix}-6\\-3\end{pmatrix}\), and \(\overrightarrow{AC}=\begin{pmatrix}-9\\1\end{pmatrix}\). 2. Use one point on each line as the position vector: \(g_{AB}:\vec{x}=\begin{pmatrix}5\\-1\end{pmatrix}+r\begin{pmatrix}-3\\4\end{pmatrix}\) \(g_{BC}:\vec{x}=\begin{pmatrix}2\\3\end{pmatrix}+s\begin{pmatrix}-6\\-3\end{pmatrix}\) \(g_{AC}:\vec{x}=\begin{pmatrix}5\\-1\end{pmatrix}+t\begin{pmatrix}-9\\1\end{pmatrix}\) Equivalent equations using another point on the same line or a nonzero scalar multiple of a direction vector are also correct.

One possible set of equations is: \(g_{AB}:\vec{x}=\begin{pmatrix}5\\-1\end{pmatrix}+r\begin{pmatrix}-3\\4\end{pmatrix}\) \(g_{BC}:\vec{x}=\begin{pmatrix}2\\3\end{pmatrix}+s\begin{pmatrix}-6\\-3\end{pmatrix}\) \(g_{AC}:\vec{x}=\begin{pmatrix}5\\-1\end{pmatrix}+t\begin{pmatrix}-9\\1\end{pmatrix}\) where \(r,s,t\in\mathbb{R}\).

- How can you find a direction vector from two points? - What must be true if a point lies on a parametric line? - The same parameter value must satisfy all three coordinate equations. - Can you use one known coordinate of \(S\) to find the parameter first?

1. A direction vector is \(\overrightarrow{PQ}=\begin{pmatrix}4\\-4\\8\end{pmatrix}\), so \(g\) can be written as \(\vec{x}=\begin{pmatrix}3\\0\\-2\end{pmatrix}+k\begin{pmatrix}4\\-4\\8\end{pmatrix}\). 2. For \(R\), the coordinate equations give \(1=3+4k\), \(2=-4k\), and \(-6=-2+8k\). Each equation gives \(k=-\frac{1}{2}\), so \(R\) lies on \(g\). 3. For \(S\), \(5=3+4k\), so \(k=\frac{1}{2}\). The \(z\)-coordinate checks because \(2=-2+8\cdot\frac{1}{2}\). Therefore, \(y=-4\cdot\frac{1}{2}=-2\).

a) \(g:\vec{x}=\begin{pmatrix}3\\0\\-2\end{pmatrix}+k\begin{pmatrix}4\\-4\\8\end{pmatrix}\), where \(k\in\mathbb{R}\) b) Yes. Point \(R\) lies on line \(g\) when \(k=-\frac{1}{2}\). c) \(y=-2\)

- Which components of each direction vector are zero, and what does that imply? - Does the position vector lie on an axis or in a coordinate plane? - What follows when the same coordinate is zero in both the position and direction vectors? - Distinguish between being an axis and being parallel to an axis.

1. For \(g\), both \(x\) and \(y\) are always \(0\), while \(z\) varies. Therefore, \(g\) is the \(z\)-axis. 2. For \(h\), the \(z\)-component of both the position vector and direction vector is \(0\). Therefore, every point on \(h\) lies in the \(xy\)-plane. 3. For \(i\), the \(x\)-component of both the position vector and direction vector is \(0\). Therefore, every point on \(i\) lies in the \(yz\)-plane. 4. For \(j\), only the \(y\)-component changes, so the line is parallel to the \(y\)-axis. Because its fixed coordinates are \(x=1\) and \(z=3\), it is not the \(y\)-axis itself.

a) \(g\) is the \(z\)-axis. b) \(h\) lies in the \(xy\)-plane. c) \(i\) lies in the \(yz\)-plane. d) \(j\) is parallel to the \(y\)-axis.

- How can you find a direction vector from two points? - How is the slope related to the components of the direction vector? - How can you find the \(y\)-intercept from the slope and one point? - What function value identifies an \(x\)-intercept?

1. A direction vector is \(\overrightarrow{PQ}=\begin{pmatrix}6\\-3\end{pmatrix}\). Thus \(h\) can be written as \(\vec{x}=\begin{pmatrix}-4\\1\end{pmatrix}+t\begin{pmatrix}6\\-3\end{pmatrix}\). 2. The slope is \(m=\frac{-3}{6}=-\frac{1}{2}\). Substitute \((-4, 1)\) into \(y=-\frac{1}{2}x+b\): \(1=-\frac{1}{2}\cdot(-4)+b\), so \(b=-1\). Therefore, \(f(x)=-\frac{1}{2}x-1\). 3. Set \(f(x)=0\): \(0=-\frac{1}{2}x-1\), so \(x=-2\).

1. \(h:\vec{x}=\begin{pmatrix}-4\\1\end{pmatrix}+t\begin{pmatrix}6\\-3\end{pmatrix}\) 2. \(f(x)=-\frac{1}{2}x-1\) 3. The \(x\)-intercept is \((-2, 0)\).

- How do you describe a line using a point and a direction vector? - How can you test whether a point satisfies a parametric equation? - How can a specified coordinate be used to find the parameter value? - The same parameter value must work in all three coordinate equations.

1. A direction vector is \(\overrightarrow{AB}=\begin{pmatrix}3\\-2\\6\end{pmatrix}\), so \(g\) can be written as \(\vec{x}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+r\begin{pmatrix}3\\-2\\6\end{pmatrix}\). 2. For \(R\), the \(x\)-coordinate gives \(7=1+3r\), so \(r=2\). Then \(2-2\cdot2=-2\) and \(-1+6\cdot2=11\). The same parameter value works in all coordinates, so \(R\) lies on \(g\). 3. Set the \(x\)-coordinate equal to \(-2\): \(1+3r=-2\), so \(r=-1\). Then \(y=2-2\cdot(-1)=4\) and \(z=-1+6\cdot(-1)=-7\). Thus \(S=(-2, 4, -7)\).

1. \(g:\vec{x}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+r\begin{pmatrix}3\\-2\\6\end{pmatrix}\) 2. Yes. Point \(R\) lies on line \(g\) when \(r=2\). 3. \(S=(-2, 4, -7)\)

- How can you find a direction vector from two given points? - What roles do the position vector and direction vector play in a parametric equation? - What must be true about the parameter if a point lies on the line? - Must the same parameter value work in all three coordinate equations?

1. A direction vector is \(\overrightarrow{AB}=\begin{pmatrix}-2\\3\\-6\end{pmatrix}\), so \(g\) can be written as \(\vec{x}=\begin{pmatrix}3\\-1\\2\end{pmatrix}+r\begin{pmatrix}-2\\3\\-6\end{pmatrix}\). 2. For \(P\), the \(x\)-coordinate gives \(5=3-2r\), so \(r=-1\). The \(y\)-coordinate checks because \(-1+3\cdot(-1)=-4\), but the \(z\)-coordinate gives \(2-6\cdot(-1)=8\ne7\). Therefore, \(P\) does not lie on \(g\).

1. \(g:\vec{x}=\begin{pmatrix}3\\-1\\2\end{pmatrix}+r\begin{pmatrix}-2\\3\\-6\end{pmatrix}\) 2. Point \(P\) does not lie on line \(g\).

- First write the line equation using points \(C\) and \(D\). - Substitute the coordinates of \(Q\) into the line equation. - Which known coordinate can be used to find \(t\)? - Substitute that parameter value into the equation for the missing coordinate.

1. A direction vector is \(\overrightarrow{CD}=\begin{pmatrix}-2\\3\\3\end{pmatrix}\), so \(h\) can be written as \(\vec{x}=\begin{pmatrix}4\\0\\-2\end{pmatrix}+t\begin{pmatrix}-2\\3\\3\end{pmatrix}\). 2. Use the \(x\)-coordinate of \(Q\): \(0=4-2t\), so \(t=2\). The \(z\)-coordinate checks because \(-2+3\cdot2=4\). Therefore, \(y=3\cdot2=6\).

1. \(h:\vec{x}=\begin{pmatrix}4\\0\\-2\end{pmatrix}+t\begin{pmatrix}-2\\3\\3\end{pmatrix}\) 2. \(y=6\)

- How can you test whether a point satisfies a parametric plane equation? - Write one equation for each coordinate. - Which coordinate equation contains only one plane parameter? - After finding the plane parameters, use the remaining coordinate equation.

1. Equating coordinates gives \(4=1+r+2s\), \(3=2+s\), and \(a=3+2r-s\). 2. The second equation gives \(s=1\). Substituting into the first equation gives \(4=1+r+2\), so \(r=1\). 3. Substitute \(r=1\) and \(s=1\) into the third equation: \(a=3+2\cdot1-1=4\).

\(a=4\)

- What two components are needed for a parametric equation of a line? - What does parallelism imply about direction vectors? - How can you form a direction vector from points \(B\) and \(C\)? - Does scaling a direction vector by a nonzero number change the line direction?

1. A direction vector for the line through \(B\) and \(C\) is \(\overrightarrow{BC}=\begin{pmatrix}-3\\3\\-3\end{pmatrix}\). 2. A parallel line can use this direction vector or any nonzero scalar multiple of it. 3. Using \(A\) as the position vector gives \(g:\vec{x}=\begin{pmatrix}-2\\5\\1\end{pmatrix}+r\begin{pmatrix}-3\\3\\-3\end{pmatrix}\).

\(g:\vec{x}=\begin{pmatrix}-2\\5\\1\end{pmatrix}+r\begin{pmatrix}-3\\3\\-3\end{pmatrix}\), where \(r\in\mathbb{R}\)

- How do you find the midpoint of two points in three dimensions? - Which part of line \(k\) determines the direction of a parallel line? - How is midpoint \(M\) used in the new line equation?

1. The midpoint of \(\overline{QR}\) is \(M=\left(\frac{0+4}{2}, \frac{2+(-2)}{2}, \frac{-3+5}{2}\right)=(2, 0, 1)\). 2. Because \(m\) is parallel to \(k\), use the direction vector \(\begin{pmatrix}-1\\1\\-2\end{pmatrix}\). 3. Therefore, \(m:\vec{x}=\begin{pmatrix}2\\0\\1\end{pmatrix}+s\begin{pmatrix}-1\\1\\-2\end{pmatrix}\).

\(m:\vec{x}=\begin{pmatrix}2\\0\\1\end{pmatrix}+s\begin{pmatrix}-1\\1\\-2\end{pmatrix}\), where \(s\in\mathbb{R}\)

- What is special about a line that passes through the origin? - How can you test whether a point satisfies the line equation? - How can the known \(y\)-coordinate be used to find the parameter?

1. Since the line passes through the origin, use the zero vector as the position vector and \(\overrightarrow{OP}=\begin{pmatrix}-2\\4\\1\end{pmatrix}\) as a direction vector. Thus \(g:\vec{x}=r\begin{pmatrix}-2\\4\\1\end{pmatrix}\). 2. For \(Q\), the coordinate equations are \(-2r=6\), \(4r=-12\), and \(r=-3\). All three give \(r=-3\), so \(Q\) lies on \(g\). 3. Set the \(y\)-coordinate equal to \(10\): \(4r=10\), so \(r=\frac{5}{2}\). Then \(x=-2\cdot\frac{5}{2}=-5\) and \(z=\frac{5}{2}\). Therefore, \(R=\left(-5, 10, \frac{5}{2}\right)\).

a) \(g:\vec{x}=r\begin{pmatrix}-2\\4\\1\end{pmatrix}\) b) Yes. Point \(Q\) lies on \(g\) when \(r=-3\). c) \(R=\left(-5, 10, \frac{5}{2}\right)\)

- How do the coordinates of a point relate to the coordinate equation? - What happens to slope when a line is vertical? - A point on a line through the origin gives a direction vector. - For the parametric form, in what direction must you move from the origin?

1. For part a, choose \(x=4\). Then \(3\cdot4-4y=0\), so \(y=3\). Thus \(P=(4, 3)\), and one parametric equation is \(\vec{x}=t\begin{pmatrix}4\\3\end{pmatrix}\). 2. For part b, substitute \((-2, 5)\) into \(y=mx\): \(5=-2m\), so \(m=-\frac{5}{2}\). Therefore, \(y=-\frac{5}{2}x\). 3. The equation \(x=0\) describes the \(y\)-axis, which is vertical and has undefined slope. No finite value of \(m\) makes \(y=mx\) represent every point with \(x=0\). One parametric equation is \(\vec{x}=t\begin{pmatrix}0\\1\end{pmatrix}\).

a) One choice is \(P=(4, 3)\), giving \(\vec{x}=t\begin{pmatrix}4\\3\end{pmatrix}\). b) \(y=-\frac{5}{2}x\) c) The line is vertical, so its slope is undefined. One parametric equation is \(\vec{x}=t\begin{pmatrix}0\\1\end{pmatrix}\).

- How do the starting point and direction vector form a parametric equation? - How can you test whether a point is reached by the parametric equation? - What parameter value must work in all three coordinate equations? - Why is the restriction \(r\ge0\) important for a ray?

1. Use the starting point as the position vector and \(\vec{u}\) as the direction vector: \(g:\vec{x}=\begin{pmatrix}2\\1\\5\end{pmatrix}+r\begin{pmatrix}-1\\2\\1\end{pmatrix}\), where \(r\ge0\). 2. For \(S\), the coordinate equations are \(-1=2-r\), \(7=1+2r\), and \(8=5+r\). Each equation gives \(r=3\). Since \(3\ge0\), the point is on the ray, so the light reaches the sensor.

a) \(g:\vec{x}=\begin{pmatrix}2\\1\\5\end{pmatrix}+r\begin{pmatrix}-1\\2\\1\end{pmatrix}\), where \(r\ge0\) b) Yes. The ray reaches the sensor when \(r=3\).

- How can you find the vector from \(P\) to \(Q\)? - What roles do the position vector and direction vector play? - How can the completed equation generate more points on the line? - Which parameter values return the original points?

1. A direction vector is \(\overrightarrow{PQ}=\begin{pmatrix}-4\\8\\-4\end{pmatrix}\). Therefore, \(g\) can be written as \(\vec{x}=\begin{pmatrix}7\\-3\\2\end{pmatrix}+r\begin{pmatrix}-4\\8\\-4\end{pmatrix}\). 2. Choose a parameter value other than \(0\) or \(1\), such as \(r=2\). Then \(S=\begin{pmatrix}7\\-3\\2\end{pmatrix}+2\begin{pmatrix}-4\\8\\-4\end{pmatrix}=\begin{pmatrix}-1\\13\\-6\end{pmatrix}\).

a) \(g:\vec{x}=\begin{pmatrix}7\\-3\\2\end{pmatrix}+r\begin{pmatrix}-4\\8\\-4\end{pmatrix}\) b) One possible point is \(S=(-1, 13, -6)\).

- First write the equation of the line through \(A\) and \(B\). - How can you test whether a point satisfies a parametric equation? - What does it mean when all three coordinate equations give the same parameter value?

1. A direction vector is \(\overrightarrow{AB}=\begin{pmatrix}3\\-3\\3\end{pmatrix}\), so \(h\) can be written as \(\vec{x}=\begin{pmatrix}1\\4\\-2\end{pmatrix}+r\begin{pmatrix}3\\-3\\3\end{pmatrix}\). 2. For \(C\), the coordinate equations are \(-2=1+3r\), \(7=4-3r\), and \(-5=-2+3r\). Each equation gives \(r=-1\). Therefore, \(C\) lies on \(h\).

Point \(C(-2, 7, -5)\) lies on line \(h\), with parameter value \(r=-1\).

- What is the general form of a line when a point and direction vector are known? - What role does the parameter play in the line equation? - How do you express a displacement from a point using vector addition? - What does a negative scalar do to the direction of a displacement?

1. The line is \(g:\vec{x}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+r\begin{pmatrix}4\\0\\-3\end{pmatrix}\). 2. For \(Q\), use \(r=\frac{5}{2}\): \(Q=\begin{pmatrix}2\\-1\\5\end{pmatrix}+\frac{5}{2}\begin{pmatrix}4\\0\\-3\end{pmatrix}=\begin{pmatrix}12\\-1\\-\frac{5}{2}\end{pmatrix}\). 3. For \(R\), use \(r=-2\): \(R=\begin{pmatrix}2\\-1\\5\end{pmatrix}-2\begin{pmatrix}4\\0\\-3\end{pmatrix}=\begin{pmatrix}-6\\-1\\11\end{pmatrix}\).

a) \(g:\vec{x}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+r\begin{pmatrix}4\\0\\-3\end{pmatrix}\), where \(r\in\mathbb{R}\) b) \(Q=\left(12, -1, -\frac{5}{2}\right)\) and \(R=(-6, -1, 11)\)

- How do a position vector and direction vector form a line equation? - What happens when a scalar multiple of the direction vector is added to the position vector? - How can a known coordinate be used to find the parameter? - Write the vector equation as separate equations for \(x\), \(y\), and \(z\).

1. The line is \(h:\vec{x}=\begin{pmatrix}0\\3\\-2\end{pmatrix}+t\begin{pmatrix}1\\-1\\2\end{pmatrix}\). 2. For \(t=-3\), \(P=\begin{pmatrix}0\\3\\-2\end{pmatrix}-3\begin{pmatrix}1\\-1\\2\end{pmatrix}=\begin{pmatrix}-3\\6\\-8\end{pmatrix}\). 3. The \(y\)-coordinate is \(3-t\). Set \(3-t=0\), so \(t=3\). Then \(x=3\) and \(z=-2+2\cdot3=4\). Therefore, \(B=(3, 0, 4)\).

a) \(h:\vec{x}=\begin{pmatrix}0\\3\\-2\end{pmatrix}+t\begin{pmatrix}1\\-1\\2\end{pmatrix}\), where \(t\in\mathbb{R}\) b) \(P=(-3, 6, -8)\) c) \(t=3\) and \(B=(3, 0, 4)\)

- How can you find a direction vector from two points? - What role does the position vector play in a parametric equation? - How can you test whether a point satisfies the line equation? - What conclusion follows if the coordinate equations give different parameter values?

1. A direction vector is \(\overrightarrow{PQ}=\begin{pmatrix}4\\-2\\4\end{pmatrix}\). Therefore, \(h:\vec{x}=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}4\\-2\\4\end{pmatrix}\). 2. Substituting \(Q\) gives \(5=1+4s\), \(0=2-2s\), and \(7=3+4s\). Each equation gives \(s=1\). 3. For \(R\), the coordinate equations give \(11=1+4s\), \(-3=2-2s\), and \(13=3+4s\). Each equation gives \(s=\frac{5}{2}\), so \(R\) lies on \(h\).

a) \(h:\vec{x}=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}4\\-2\\4\end{pmatrix}\) b) \(s=1\) c) Yes. Point \(R\) lies on \(h\) when \(s=\frac{5}{2}\).

- How can a slope be converted into a direction vector? - What does an equation with only one variable, such as \(x=7\), say about every point on the line? - Can you find two points that satisfy each equation and subtract their coordinates? - Choose a point on the line that is easy to calculate.

1. For \(g_1\), the point \((0, -2)\) lies on the line, and slope \(\frac{3}{4}\) gives direction vector \(\begin{pmatrix}4\\3\end{pmatrix}\). Thus \(\vec{x}=\begin{pmatrix}0\\-2\end{pmatrix}+r\begin{pmatrix}4\\3\end{pmatrix}\). 2. For \(g_2\), two convenient points are \((2, 0)\) and \((0, -5)\). A direction vector is \(\begin{pmatrix}2\\5\end{pmatrix}\), so \(\vec{x}=\begin{pmatrix}2\\0\end{pmatrix}+s\begin{pmatrix}2\\5\end{pmatrix}\). 3. For \(g_3\), the \(x\)-coordinate is always \(7\), so the line is vertical. Use point \((7, 0)\) and direction vector \(\begin{pmatrix}0\\1\end{pmatrix}\): \(\vec{x}=\begin{pmatrix}7\\0\end{pmatrix}+t\begin{pmatrix}0\\1\end{pmatrix}\).

a) \(g_1:\vec{x}=\begin{pmatrix}0\\-2\end{pmatrix}+r\begin{pmatrix}4\\3\end{pmatrix}\) b) \(g_2:\vec{x}=\begin{pmatrix}2\\0\end{pmatrix}+s\begin{pmatrix}2\\5\end{pmatrix}\) c) \(g_3:\vec{x}=\begin{pmatrix}7\\0\end{pmatrix}+t\begin{pmatrix}0\\1\end{pmatrix}\)

- How can you compare a point's coordinates with the parametric equation? - What must be true about the parameter for a point to lie on the line? - After finding the parameter, how can you test whether it lies in a specified interval? - Check every coordinate equation.

1. For \(P\), the coordinate equations all give \(r=2\), so \(P\) lies on \(g\). Since \(2\in[0, 3]\), it also lies on the specified portion. 2. For \(Q\), the coordinate equations all give \(r=-2\), so \(Q\) lies on \(g\). Since \(-2\notin[0, 3]\), it does not lie on the specified portion. 3. For \(R\), the \(x\)-coordinate gives \(r=-\frac{1}{2}\), while the \(y\)-coordinate gives \(r=\frac{1}{2}\). No single parameter value works, so \(R\) does not lie on \(g\).

a) Points \(P\) and \(Q\) lie on \(g\); point \(R\) does not. b) Only \(P\) lies on the portion defined by \(0\le r\le3\).

- Which coordinate is always \(0\) in the \(xy\)-plane? - How can the condition \(x=z\) be written as an equation in \(t\)? - Give all three coordinates of each point.

1. For \(C\), a point in the \(xy\)-plane has \(z=0\). Solve \(2+t=0\), so \(t=-2\). Then \(x=1-(-2)=3\) and \(y=4+2\cdot(-2)=0\). Thus \(C=(3, 0, 0)\). 2. For \(D\), require \(x=z\): \(1-t=2+t\). Then \(t=-\frac{1}{2}\). Substitution gives \(x=\frac{3}{2}\), \(y=3\), and \(z=\frac{3}{2}\). Thus \(D=\left(\frac{3}{2},3,\frac{3}{2}\right)\).

a) \(C=(3, 0, 0)\) b) \(D=\left(\frac{3}{2},3,\frac{3}{2}\right)\)

- How can you parametrize the line through two points? - What parameter interval represents the segment between the endpoints? - What happens when the parameter is less than \(0\) or greater than \(1\)? - The same parameter value must work for all three coordinates.

1. Parametrize the line through \(A\) and \(B\): \(\vec{x}=\begin{pmatrix}5\\-1\\2\end{pmatrix}+k\begin{pmatrix}-6\\6\\9\end{pmatrix}\). The segment corresponds to \(0\le k\le1\). 2. For \(P\), all three coordinate equations give \(k=\frac{2}{3}\). Since \(\frac{2}{3}\in[0, 1]\), \(P\) lies on the segment. 3. For \(Q\), all three coordinate equations give \(k=-\frac{1}{3}\). Thus \(Q\) lies on the line but not on the segment. 4. For \(R\), all three coordinate equations give \(k=\frac{1}{3}\). Since \(\frac{1}{3}\in[0, 1]\), \(R\) lies on the segment.

Points \(P\) and \(R\) lie on \(\overline{AB}\). Point \(Q\) lies on the line through \(A\) and \(B\), but not on the segment.

- How do you find points when their parameter values are given? - What parameter values represent points before, between, and beyond the endpoints? - For \(N\), first test membership on the full line, then check the parameter interval.

1. At \(r=0\), \(C=(-2, 4, 0)\). At \(r=1\), \(D=(4, -2, 6)\). 2. For \(M\), the \(x\)-coordinate gives \(-2+6r=1\), so \(r=\frac{1}{2}\). The other coordinates check: \(4-6\cdot\frac{1}{2}=1\) and \(6\cdot\frac{1}{2}=3\). Since \(\frac{1}{2}\in[0, 1]\), \(M\) lies on the segment. 3. For \(N\), the \(x\)-coordinate gives \(-2+6r=6\), so \(r=\frac{4}{3}\). The other coordinates check: \(4-6\cdot\frac{4}{3}=-4\) and \(6\cdot\frac{4}{3}=8\). Thus \(N\) lies on \(g\), but \(\frac{4}{3}>1\), so it is outside the segment.

a) \(C=(-2, 4, 0)\) and \(D=(4, -2, 6)\) b) Yes. Point \(M\) lies on the segment when \(r=\frac{1}{2}\). c) Point \(N\) lies on the line when \(r=\frac{4}{3}\), but it is not on the segment.

- First write a parametric equation for every point the laser can reach. - How can the known \(x\)-coordinate be used to find the parameter? - Use the same parameter value in the other coordinate equations. - Check that the parameter is consistent with the direction of the ray.

1. The laser ray is \(\vec{x}=\begin{pmatrix}5\\-2\\1.2\end{pmatrix}+t\begin{pmatrix}3\\4\\-0.1\end{pmatrix}\), where \(t\ge0\). 2. Use the known \(x\)-coordinate: \(5+3t=17\), so \(t=4\). 3. Then \(y=-2+4\cdot4=14\) and \(z=1.2+4\cdot(-0.1)=0.8\). Since \(t=4\ge0\), the point lies on the ray.

\(y=14\) and \(z=0.8\), so the center is \((17, 14, 0.8)\).

- How are a line's normal vector and direction vector related? - Which vector can be read from the coefficients of \(4x+3y=12\)? - How can you find a vector perpendicular to \(\begin{pmatrix}a\\b\end{pmatrix}\)? - Parallel lines can use the same direction vector.

1. A normal vector to \(g\) is \(\begin{pmatrix}4\\3\end{pmatrix}\). 2. A perpendicular direction vector is \(\begin{pmatrix}3\\-4\end{pmatrix}\), since \(4\cdot3+3\cdot(-4)=0\). 3. A parallel line can use this direction vector. Using \(P\) as the position vector gives \(h:\vec{x}=\begin{pmatrix}1\\5\end{pmatrix}+\lambda\begin{pmatrix}3\\-4\end{pmatrix}\).

\(h:\vec{x}=\begin{pmatrix}1\\5\end{pmatrix}+\lambda\begin{pmatrix}3\\-4\end{pmatrix}\), where \(\lambda\in\mathbb{R}\)

- What does lying in the \(xz\)-plane imply about the \(y\)-coordinate? - How can you choose values satisfying \(5x=2z\)? - What position vector is used for a line through the origin? - How can you test whether a point satisfies the parametric equation?

1. Because the line lies in the \(xz\)-plane, every point has \(y=0\). The equation \(5x-2z=0\) gives \(5x=2z\). Choosing \(x=2\) gives \(z=5\), so one direction vector is \(\begin{pmatrix}2\\0\\5\end{pmatrix}\). Since the line passes through the origin, \(h:\vec{x}=s\begin{pmatrix}2\\0\\5\end{pmatrix}\). 2. For \(Q\), the \(x\)-coordinate gives \(4=2s\), so \(s=2\). Then \(y=0\) and \(z=5\cdot2=10\). Therefore, \(Q\) lies on \(h\).

1. One possible direction vector is \(\begin{pmatrix}2\\0\\5\end{pmatrix}\), and \(h:\vec{x}=s\begin{pmatrix}2\\0\\5\end{pmatrix}\). 2. Yes. Point \(Q\) lies on \(h\) when \(s=2\).

- How can you test whether a point satisfies a parametric line equation? - What equation does the given \(x\)-coordinate create for \(\lambda\)? - What direction vector points along the \(y\)-axis? - A parametric line equation uses one point and one direction vector.

1. For \(P\), the first coordinate gives \(1=3-\lambda\), so \(\lambda=2\). The other coordinates also satisfy the equation: \(2=-2+2\cdot2\) and \(15=5+5\cdot2\). Therefore, \(P\) lies on \(g\). 2. For \(Q\), solve \(5=3-\lambda\), which gives \(\lambda=-2\). Then \(y=-2+2\cdot(-2)=-6\) and \(z=5+5\cdot(-2)=-5\). Thus \(Q=(5, -6, -5)\). 3. A direction vector parallel to the \(y\)-axis is \(\begin{pmatrix}0\\1\\0\end{pmatrix}\). Using \(P\) as the position vector gives \(h:\vec{x}=\begin{pmatrix}1\\2\\15\end{pmatrix}+\mu\begin{pmatrix}0\\1\\0\end{pmatrix}\).

a) Point \(P\) lies on \(g\) when \(\lambda=2\). b) \(Q=(5, -6, -5)\) c) \(h:\vec{x}=\begin{pmatrix}1\\2\\15\end{pmatrix}+\mu\begin{pmatrix}0\\1\\0\end{pmatrix}\), where \(\mu\in\mathbb{R}\)

- How do you find a direction vector from two points? - What must be true about the parameter in all three coordinate equations? - What do direction vectors of parallel lines have in common? - What position vector represents the origin?

1. A direction vector from \(C\) to \(D\) is \(\overrightarrow{CD}=\begin{pmatrix}4-2\\1-3\\3-(-1)\end{pmatrix}=\begin{pmatrix}2\\-2\\4\end{pmatrix}\). Therefore, \(k:\vec{x}=\begin{pmatrix}2\\3\\-1\end{pmatrix}+s\begin{pmatrix}2\\-2\\4\end{pmatrix}\). 2. For \(E\), the first coordinate gives \(1=2+2s\), so \(s=-\frac{1}{2}\). This value also gives \(3-2\cdot\left(-\frac{1}{2}\right)=4\) and \(-1+4\cdot\left(-\frac{1}{2}\right)=-3\). Therefore, \(E\) lies on \(k\). 3. A line parallel to \(k\) can use the same direction vector. Since \(m\) passes through the origin, \(m:\vec{x}=t\begin{pmatrix}2\\-2\\4\end{pmatrix}\).

a) \(k:\vec{x}=\begin{pmatrix}2\\3\\-1\end{pmatrix}+s\begin{pmatrix}2\\-2\\4\end{pmatrix}\) b) Yes. Point \(E\) lies on \(k\) when \(s=-\frac{1}{2}\). c) \(m:\vec{x}=t\begin{pmatrix}2\\-2\\4\end{pmatrix}\), where \(t\in\mathbb{R}\)

- Substitute the given values for both parameters. - Multiply each direction vector by its scalar. - Add the vectors component by component. - The components of the resulting position vector are the point's coordinates.

1. For \(r=2\) and \(s=-1\), \(\vec{x}=\begin{pmatrix}1\\-2\\4\end{pmatrix}+2\begin{pmatrix}1\\3\\-2\end{pmatrix}-\begin{pmatrix}0\\1\\5\end{pmatrix}=\begin{pmatrix}3\\3\\-5\end{pmatrix}\). 2. For \(r=-3\) and \(s=\frac{1}{2}\), \(\vec{x}=\begin{pmatrix}1\\-2\\4\end{pmatrix}-3\begin{pmatrix}1\\3\\-2\end{pmatrix}+\frac{1}{2}\begin{pmatrix}0\\1\\5\end{pmatrix}=\begin{pmatrix}-2\\-\frac{21}{2}\\\frac{25}{2}\end{pmatrix}\).

a) \((3, 3, -5)\) b) \(\left(-2,-\frac{21}{2},\frac{25}{2}\right)\)

- Use point \(A\) as the position vector in the plane equation. - Substitute the given parameter values. - Multiply each direction vector by its parameter. - Add corresponding components carefully, especially with negative signs.

1. A parametric equation of the plane is \(\vec{x}=\begin{pmatrix}2\\0\\3\end{pmatrix}+k\begin{pmatrix}1\\1\\1\end{pmatrix}+m\begin{pmatrix}-1\\2\\0\end{pmatrix}\). 2. For \(k=4\) and \(m=-2\), \(\vec{p}=\begin{pmatrix}2\\0\\3\end{pmatrix}+4\begin{pmatrix}1\\1\\1\end{pmatrix}-2\begin{pmatrix}-1\\2\\0\end{pmatrix}=\begin{pmatrix}8\\0\\7\end{pmatrix}\). 3. For \(k=\frac{3}{2}\) and \(m=\frac{3}{2}\), \(\vec{p}=\begin{pmatrix}2\\0\\3\end{pmatrix}+\frac{3}{2}\begin{pmatrix}1\\1\\1\end{pmatrix}+\frac{3}{2}\begin{pmatrix}-1\\2\\0\end{pmatrix}=\begin{pmatrix}2\\\frac{9}{2}\\\frac{9}{2}\end{pmatrix}\).

a) \(\vec{p}=\begin{pmatrix}8\\0\\7\end{pmatrix}\) b) \(\vec{p}=\begin{pmatrix}2\\\frac{9}{2}\\\frac{9}{2}\end{pmatrix}\)

- Three points determine a unique plane when they are not collinear. - Compare two connecting vectors for scalar-multiple relationships. - Test whether the point lies on the given line. - A point already on a line adds no new direction for defining a plane.

1. For part a, \(\overrightarrow{AB}=\begin{pmatrix}2\\3\\0\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}4\\6\\1\end{pmatrix}\) are not scalar multiples because their third coordinates differ in a way no scalar multiple can match. Thus the points are not collinear and determine a unique plane. 2. For part b, substituting \(s=2\) into the line gives \(P=(5, 0, 7)\). Thus \(P\) lies on \(g\). A line together with a point already on it does not determine a unique plane.

a) Yes. The three points are not collinear. b) No. The point \(P\) lies on \(g\), so infinitely many planes contain the given line and point.

- Equate corresponding coordinates of the two line equations. - First solve the equations that do not contain \(k\). - Use the remaining coordinate equation to find \(k\).

1. Equating coordinates gives \(2+r=4+2s\), \(1+r=3+s\), and \(2r=k\). 2. The first two equations simplify to \(r-2s=2\) and \(r-s=2\). Subtracting gives \(s=0\), and then \(r=2\). 3. The third equation gives \(k=2r=4\). 4. Substituting \(r=2\) into \(g\) gives \(S=(4, 3, 4)\).

\(k=4\), and the lines intersect at \(S=(4, 3, 4)\).

- First decide on a point that the line must contain. - Which direction-vector components must be zero for a line to be parallel to a coordinate axis or plane? - What condition must every point satisfy for the line to lie in a specified coordinate plane? - Each part has infinitely many correct equations.

1. For the \(x\)-axis, use the origin and a direction vector parallel to the \(x\)-axis: \(\vec{x}=\lambda\begin{pmatrix}1\\0\\0\end{pmatrix}\). 2. A line parallel to the \(z\)-axis has a direction vector with only a nonzero \(z\)-component. Through \(P\), one choice is \(\vec{x}=\begin{pmatrix}2\\5\\0\end{pmatrix}+\lambda\begin{pmatrix}0\\0\\1\end{pmatrix}\). 3. To remain in the \(xz\)-plane, both \(y\)-components must be \(0\). Choose a point and direction so the line misses the origin and changes in both \(x\) and \(z\): \(\vec{x}=\begin{pmatrix}1\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\1\end{pmatrix}\). 4. A line parallel to the \(xy\)-plane has direction-vector \(z\)-component \(0\). At height \(z=4\), one choice is \(\vec{x}=\begin{pmatrix}0\\0\\4\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\0\end{pmatrix}\).

a) \(k:\vec{x}=\lambda\begin{pmatrix}1\\0\\0\end{pmatrix}\) b) \(k:\vec{x}=\begin{pmatrix}2\\5\\0\end{pmatrix}+\lambda\begin{pmatrix}0\\0\\1\end{pmatrix}\) c) \(k:\vec{x}=\begin{pmatrix}1\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\1\end{pmatrix}\) d) \(k:\vec{x}=\begin{pmatrix}0\\0\\4\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\0\end{pmatrix}\)

- Which coordinate must equal \(0\) for a point to lie in the \(xy\)-plane? - How can you find a point on the line from a parameter value? - How are different direction vectors for the same line related? - Use the magnitude formula for a vector.

1. A point in the \(xy\)-plane has \(z=0\). Set the \(z\)-coordinate equal to \(0\): \(1-6r=0\), so \(r=\frac{1}{6}\). 2. Substitute \(r=\frac{1}{6}\) into the original equation. This gives the point \(P\left(\frac{16}{3},-\frac{4}{3},0\right)\). 3. The original direction vector has magnitude \(\sqrt{2^2+4^2+(-6)^2}=\sqrt{56}=2\sqrt{14}\). Dividing it by \(2\) gives \(\begin{pmatrix}1\\2\\-3\end{pmatrix}\), whose magnitude is \(\sqrt{14}\). 4. Therefore, one valid equation is \(\vec{x}=\begin{pmatrix}\frac{16}{3}\\-\frac{4}{3}\\0\end{pmatrix}+s\begin{pmatrix}1\\2\\-3\end{pmatrix}\).

One possible equation is \(g:\vec{x}=\begin{pmatrix}\frac{16}{3}\\-\frac{4}{3}\\0\end{pmatrix}+s\begin{pmatrix}1\\2\\-3\end{pmatrix}\), where \(s\in\mathbb{R}\). The opposite direction vector is also valid.

- Equate the two position vectors coordinate by coordinate. - Begin with the coordinate equation containing only one line parameter. - Use the next equation to find the other line parameter. - Then determine \(b\) and calculate the common point.

1. Equating coordinates gives \(2+r=4\), \(3=b+s\), and \(1+2r=3+s\). 2. The first equation gives \(r=2\). The third then gives \(5=3+s\), so \(s=2\). 3. The second equation gives \(3=b+2\), so \(b=1\). 4. Substituting \(r=2\) into \(g\) gives the intersection point \((4, 3, 5)\).

\(b=1\), and the lines intersect at \((4, 3, 5)\).

- Which points correspond to \(\lambda=0\) and \(\lambda=1\)? - Which parameter value gives the midpoint of \(\overline{AB}\)? - What happens to the point when the parameter is negative? - Think of \(\overrightarrow{AB}\) as a step from \(A\): where does a half-step or a backward step land?

1. The direction vector is \(\overrightarrow{AB}=\begin{pmatrix}3\\6\\-3\end{pmatrix}\). 2. For \(\lambda=0.4\), \(\vec{x}=\begin{pmatrix}1\\-2\\3\end{pmatrix}+0.4\begin{pmatrix}3\\6\\-3\end{pmatrix}=\begin{pmatrix}2.2\\0.4\\1.8\end{pmatrix}\). Thus \(P=(2.2, 0.4, 1.8)\). 3. A point lies strictly between \(A\) and \(B\) when \(0<\lambda<1\). Its distance from \(A\) is \(\lambda\lVert\overrightarrow{AB}\rVert\), and its distance from \(B\) is \((1-\lambda)\lVert\overrightarrow{AB}\rVert\). The first is greater when \(\lambda>\frac{1}{2}\). Therefore, \(\frac{1}{2}<\lambda<1\). 4. When \(\lambda<0\), the point lies outside segment \(\overline{AB}\), beyond \(A\) on the side opposite \(B\).

a) \(P=(2.2, 0.4, 1.8)\) b) \(\lambda\in\left(\frac{1}{2},1\right)\) c) Point \(P\) lies beyond \(A\) on the extension of the line opposite \(B\).

- Which coordinate represents altitude above the ground? - How is the parameter \(t\) related to elapsed time? - How do you find the magnitude of a vector in three dimensions? - For the distance from the starting point, should you use the position vector or the displacement vector?

1. The altitude is the \(z\)-coordinate. Set \(50+5t=300\). Then \(5t=250\), so \(t=50\). The balloon reaches the altitude after \(50\,\text{s}\). 2. At \(t=50\), the displacement from the starting point is \(50\begin{pmatrix}4\\2\\5\end{pmatrix}=\begin{pmatrix}200\\100\\250\end{pmatrix}\). 3. Its magnitude is \(\sqrt{200^2+100^2+250^2}=\sqrt{112500}=150\sqrt{5}\approx335.4\). Therefore, the balloon is about \(335.4\,\text{m}\) from its starting point.

a) \(50\,\text{s}\) b) \(150\sqrt{5}\,\text{m}\approx335.4\,\text{m}\)

- What must the altitude coordinate equal when the plane reaches the ground? - How can you find the position at a given time? - Which distance formula gives the distance from a point to the origin?

1. The plane reaches the ground when its \(z\)-coordinate is \(0\). Solve \(600-20t=0\), which gives \(t=30\). Therefore, the plane reaches the ground after \(30\,\text{s}\). 2. At \(t=30\), the position is \(\vec{x}(30)=\begin{pmatrix}-2000+30\cdot80\\1500+30\cdot(-60)\\600+30\cdot(-20)\end{pmatrix}=\begin{pmatrix}400\\-300\\0\end{pmatrix}\). 3. The distance from the origin is \(\sqrt{400^2+(-300)^2}=\sqrt{250{,}000}=500\).

a) \(30\,\text{s}\) b) \(500\,\text{m}\)

- First check whether the direction vectors are scalar multiples. - Equate corresponding coordinates of the line equations. - Solve two equations for the two line parameters. - Verify the values in the third equation before finding the point. - Use the parameter value to compute the coordinates of the intersection point.

1. The direction vectors are not scalar multiples, so the lines are not parallel. 2. Equating coordinates gives \(2r+s=3\), \(r-2s=-1\), and \(-3r+4s=1\). 3. Solving the first two equations gives \(r=1\) and \(s=1\). These values satisfy the third equation because \(-3+4=1\). 4. Substituting \(r=1\) into \(g\) gives \(S=(3, 2, 1)\). Therefore, the lines intersect.

The lines intersect at \(S=(3, 2, 1)\).

- First check whether the direction vectors are scalar multiples. - If they are not parallel, the lines either intersect or are skew. - Equate corresponding coordinates to form a system. - What does it mean if values that satisfy two coordinate equations fail in the third?

1. The direction vectors are not scalar multiples, so the lines are not parallel or identical. 2. Equating coordinates gives \(\lambda-3\mu=2\), \(2\lambda+\mu=-3\), and \(2\lambda-\mu=4\). 3. Solving the first two equations gives \(\lambda=-1\) and \(\mu=-1\). 4. These values do not satisfy the third equation because \(2(-1)-(-1)=-1\neq4\). 5. The lines are nonparallel and do not intersect, so they are skew.

The lines \(g\) and \(h\) are skew.

- First determine whether the lines can ever be parallel. - Use the coordinate equations that do not contain \(a\) to find the line parameters. - What value of \(a\) makes the remaining coordinate equation true? - What is the relationship when nonparallel lines have no common point?

1. The direction vectors are not scalar multiples, so the lines are never parallel. 2. Equating coordinates gives \(1+\lambda=a+3\mu\), \(4-2\lambda=2+2\mu\), and \(3+2\lambda=1\). 3. The third equation gives \(\lambda=-1\). Substituting this into the second equation gives \(\mu=2\). 4. The first equation is then satisfied only when \(0=a+6\), so \(a=-6\). 5. For \(a=-6\), substituting \(\lambda=-1\) into \(g\) gives the intersection point \((0, 6, 1)\). For every \(a\neq-6\), the nonparallel lines do not intersect and are skew.

For \(a=-6\), the lines intersect at \((0, 6, 1)\). For \(a\neq-6\), the lines are skew.

- Check whether the three points are collinear. - A contradiction in a point-on-line test shows that the point is not on the line. - For a plane through a line and an outside point, use the line direction and a connecting vector as spanning vectors. - The two spanning vectors must be linearly independent.

1. For part a, \(\overrightarrow{AB}=\begin{pmatrix}1\\2\\1\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}3\\6\\3\end{pmatrix}=3\overrightarrow{AB}\). The three points are collinear, so they do not determine a unique plane. 2. For part b, point \(P\) is not on \(g\), because every point on \(g\) has first coordinate \(3\), while \(P\) has first coordinate \(2\). A line and a point not on it determine a unique plane. 3. Use the line direction \(\begin{pmatrix}0\\1\\1\end{pmatrix}\) and the vector from the line's given point to \(P\), \(\begin{pmatrix}-1\\1\\0\end{pmatrix}\), as spanning vectors. One equation is \(E:\vec{x}=\begin{pmatrix}3\\1\\2\end{pmatrix}+t\begin{pmatrix}0\\1\\1\end{pmatrix}+s\begin{pmatrix}-1\\1\\0\end{pmatrix}\).

a) No unique plane is determined because the three points are collinear. b) A unique plane is determined. One possible equation is \(E:\vec{x}=\begin{pmatrix}3\\1\\2\end{pmatrix}+t\begin{pmatrix}0\\1\\1\end{pmatrix}+s\begin{pmatrix}-1\\1\\0\end{pmatrix}\), where \(s,t\in\mathbb{R}\).

- Write one equation for each coordinate. - Express \(\lambda\) in terms of \(a\) using the first coordinate. - Then express \(\mu\) in terms of \(a\). - Substitute into the remaining coordinate equation.

1. Equating coordinates gives \(a+1=1+\lambda\), \(a=2-\lambda+2\mu\), and \(9=1+2\lambda+\mu\). 2. The first equation gives \(\lambda=a\). Substituting into the second equation gives \(a=2-a+2\mu\), so \(\mu=a-1\). 3. Substitute both expressions into the third equation: \(9=1+2a+(a-1)=3a\). Therefore, \(a=3\). 4. For \(a=3\), the plane parameters are \(\lambda=3\) and \(\mu=2\), which satisfy all three coordinate equations.

\(a=3\)

- Find two vectors connecting one point to the other two. - The spanning vectors must be linearly independent. - To test a point, set the parametric expression equal to the point's coordinates. - Solve the resulting system for the two parameters.

1. Two spanning vectors are \(\overrightarrow{PQ}=\begin{pmatrix}3\\-2\\-2\end{pmatrix}\) and \(\overrightarrow{PR}=\begin{pmatrix}-3\\3\\-3\end{pmatrix}\). They are not scalar multiples, so one equation is \(E:\vec{x}=\begin{pmatrix}1\\2\\3\end{pmatrix}+r\begin{pmatrix}3\\-2\\-2\end{pmatrix}+s\begin{pmatrix}-3\\3\\-3\end{pmatrix}\). 2. Setting this expression equal to \(S\) gives \(7=1+3r-3s\), \(-2=2-2r+3s\), and \(-1=3-2r-3s\). The system has the solution \(r=2\), \(s=0\). Therefore, \(S\) lies in the plane.

1. One equation is \(E:\vec{x}=\begin{pmatrix}1\\2\\3\end{pmatrix}+r\begin{pmatrix}3\\-2\\-2\end{pmatrix}+s\begin{pmatrix}-3\\3\\-3\end{pmatrix}\). 2. Yes. The parameter values \(r=2\) and \(s=0\) produce \(S\).

- A parametric plane equation uses one point and two spanning vectors. - A normal vector is perpendicular to both spanning vectors; use dot products to find one. - Expand the point-normal equation to obtain Cartesian form. - Substitute the given \(x\)- and \(y\)-coordinates to find the height \(z\).

1. The vectors \(\overrightarrow{P_1P_2}=\begin{pmatrix}40\\0\\1\end{pmatrix}\) and \(\overrightarrow{P_1P_3}=\begin{pmatrix}0\\40\\-1\end{pmatrix}\) span the plane. Thus \(E:\vec{x}=\begin{pmatrix}0\\0\\30\end{pmatrix}+r\begin{pmatrix}40\\0\\1\end{pmatrix}+s\begin{pmatrix}0\\40\\-1\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. The equations \(40a+c=0\) and \(40b-c=0\) are satisfied by \(\vec{n}=\begin{pmatrix}-1\\1\\40\end{pmatrix}\). A point-normal equation is \(\left(\vec{x}-\begin{pmatrix}0\\0\\30\end{pmatrix}\right)\cdot\begin{pmatrix}-1\\1\\40\end{pmatrix}=0\). 3. Expanding gives \(-x+y+40z=1200\). 4. At \(x=20\) and \(y=20\), \(-20+20+40z=1200\), so \(z=30\).

a) \(E:\vec{x}=\begin{pmatrix}0\\0\\30\end{pmatrix}+r\begin{pmatrix}40\\0\\1\end{pmatrix}+s\begin{pmatrix}0\\40\\-1\end{pmatrix}\) b) One normal vector is \(\vec{n}=\begin{pmatrix}-1\\1\\40\end{pmatrix}\), and one point-normal equation is \(\left(\vec{x}-\begin{pmatrix}0\\0\\30\end{pmatrix}\right)\cdot\begin{pmatrix}-1\\1\\40\end{pmatrix}=0\). c) \(E:-x+y+40z=1200\) d) \(30\,\text{in.}\)

- How can you test whether a point satisfies a parametric plane equation? - A point on the \(y\)-axis has which two coordinates equal to \(0\)? - For point \(P\), how can you use the fact that its first and third coordinates are equal? - Solve two coordinate equations for the plane parameters, then check or use the third.

1. For \(A\), equating coordinates gives \(3+\lambda+2\mu=6\), \(1+2\lambda-\mu=2\), and \(-2+\lambda+3\mu=2\). The first two equations give \(\lambda=1\) and \(\mu=1\), and these values satisfy the third equation. Therefore, \(A\) lies in \(E\). 2. For \(B\), the first two coordinate equations give \(\lambda=0\) and \(\mu=-1\). The third coordinate would then be \(-2+0+3\cdot(-1)=-5\), not \(3\). Therefore, \(B\) does not lie in \(E\). 3. For \(P\), equating the first and third coordinate expressions gives \(3+\lambda+2\mu=-2+\lambda+3\mu\), so \(\mu=5\). The second coordinate gives \(4=1+2\lambda-5\), so \(\lambda=4\). Then \(p=3+4+2\cdot5=17\). 4. A point on the \(y\)-axis has the form \((0,y,0)\). Solving \(3+\lambda+2\mu=0\) and \(-2+\lambda+3\mu=0\) gives \(\mu=5\) and \(\lambda=-13\). Thus \(y=1+2\cdot(-13)-5=-30\), so the intersection point is \((0, -30, 0)\).

a) Point \(A\) lies in \(E\); point \(B\) does not. b) \(p=17\) c) \((0, -30, 0)\)

- Substitute each point's coordinates into the parametric plane equation. - Use equal coordinates to equate the corresponding coordinate expressions. - When one coordinate is \(0\), begin with that coordinate equation. - What plane contains every point whose \(x\)-coordinate is \(0\)?

1. For \(Q\), equating coordinates gives \(q=2+r\), \(q=r+2s\), and \(5=1+2r-s\). The first two equations give \(s=1\). Then \(5=1+2r-1\), so \(r=\frac{5}{2}\) and \(q=2+\frac{5}{2}=\frac{9}{2}\). 2. For \(R\), the equations are \(k=2+r\), \(k=r+2s\), and \(k=1+2r-s\). The first two give \(s=1\). Then the third gives \(k=2r\). Equating this with \(k=2+r\) gives \(r=2\), so \(k=4\). 3. For \(T\), the first coordinate gives \(0=2+r\), so \(r=-2\). The remaining coordinates give \(t=-2+2s\) and \(t=-3-s\). Equating them gives \(s=-\frac{1}{3}\), so \(t=-\frac{8}{3}\). 4. Thus \(T=\left(0,-\frac{8}{3},-\frac{8}{3}\right)\). Because \(x=0\), it lies in the \(yz\)-plane; because \(y=z\), it lies on the line \(y=z\) in that plane.

a) \(q=\frac{9}{2}\) b) \(k=4\) c) \(t=-\frac{8}{3}\). Point \(T\) lies in the \(yz\)-plane on the line \(y=z\).

- Identify where each trace line meets a coordinate axis. - Use the plane intercept form \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\). - Clear the fractions to obtain a Cartesian equation. - Test the point by substituting its coordinates into the plane equation.

1. The common point on the \(x\)-axis is \(S_1=(5, 0, 0)\). Setting \(r=1\) gives the \(y\)-axis point \(S_2=(0, 2, 0)\), and setting \(s=1\) gives the \(z\)-axis point \(S_3=(0, 0, 10)\). 2. The intercept form is \(\frac{x}{5}+\frac{y}{2}+\frac{z}{10}=1\). Multiplying by \(10\) gives \(E:2x+5y+z=10\). 3. For \(Q\), \(2\cdot1+5\cdot1+3=10\). Therefore, \(Q\) lies in the plane.

a) \(S_1=(5, 0, 0)\), \(S_2=(0, 2, 0)\), and \(S_3=(0, 0, 10)\) b) \(E:2x+5y+z=10\) c) Yes, \(Q\) lies in \(E\).

- What region is formed when each of two independent vector coefficients ranges from \(0\) to \(1\)? - How can a dot product determine whether the sides are perpendicular? - What restriction on the sum of the coefficients selects the triangle rather than the entire parallelogram? - Substitute the given parameter values into the plane equation.

1. The plane direction vectors are \(\overrightarrow{AB}=\begin{pmatrix}4\\0\\0\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}0\\3\\0\end{pmatrix}\). 2. Allowing \(\lambda\) and \(\mu\) to vary independently from \(0\) to \(1\) forms a parallelogram. Since \(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\), the sides are perpendicular, so the figure is a rectangle. 3. Its side lengths are \(4\) and \(3\), so its area is \(4\cdot3=12\) square units. 4. Triangle \(ABC\) is described by \(\lambda\ge0\), \(\mu\ge0\), and \(\lambda+\mu\le1\). 5. For \(\lambda=1\) and \(\mu=1\), \(D=A+\overrightarrow{AB}+\overrightarrow{AC}=(1, 1, 2)+(4, 0, 0)+(0, 3, 0)=(5, 4, 2)\).

a) A rectangle with area \(12\) square units b) \(\lambda\ge0\), \(\mu\ge0\), and \(\lambda+\mu\le1\) c) \(D=(5, 4, 2)\)

- Compare the lengths and dot product of the two direction vectors. - How do the parameter intervals affect the side lengths? - What parameter values lie at the center of both intervals? - The distance from parameter \(-1\) to \(1\) is \(2\).

1. Let \(\vec{u}=\begin{pmatrix}2\\1\\0\end{pmatrix}\) and \(\vec{v}=\begin{pmatrix}-1\\2\\0\end{pmatrix}\). Their dot product is \(\vec{u}\cdot\vec{v}=2\cdot(-1)+1\cdot2+0=0\), so they are perpendicular. 2. Their lengths are equal: \(\lVert\vec{u}\rVert=\sqrt{5}\) and \(\lVert\vec{v}\rVert=\sqrt{5}\). Both parameters vary across intervals of length \(2\), so the four sides are equal and adjacent sides are perpendicular. Thus \(M\) is a square. 3. The parameter intervals are centered at \(r=0\) and \(s=0\), so the center is \((0, 0, 5)\). 4. Each side has length \(2\sqrt{5}\). Therefore, the area is \((2\sqrt{5})^2=20\) square units.

a) The direction vectors are perpendicular and have equal length, and both parameter intervals have length \(2\); therefore, \(M\) is a square. b) \((0, 0, 5)\) c) \(20\) square units

- Set corresponding coordinates of the two lines equal. - Two intersecting lines with different directions determine a plane. - Let \(\vec{n}\) be perpendicular to both line directions and use two dot-product equations to find it. - Substitute any point on the plane to find the constant term.

1. Equating corresponding coordinates gives \(r=2s\), \(1+2r=6-s\), and \(-1+2r=2+s\). Solving gives \(s=1\) and \(r=2\), so \(S=(4, 5, 3)\). 2. The direction vectors \(\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\begin{pmatrix}2\\-1\\1\end{pmatrix}\) are not scalar multiples. Since the lines intersect and have different directions, they determine a unique plane. 3. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be a normal vector. Perpendicularity to both line directions gives \(a+2b+2c=0\) and \(2a-b+c=0\). One solution is \(\vec{n}=\begin{pmatrix}4\\3\\-5\end{pmatrix}\). Using a point on the plane gives \(E:4x+3y-5z=16\).

a) \(S=(4, 5, 3)\) b) The intersecting nonparallel lines determine a unique plane, and one equation is \(E:4x+3y-5z=16\).

- Compare the direction vectors first. - For parallel lines, test whether a point from one line lies on the other. - Use the common direction and a vector connecting the lines as spanning vectors. - Find a normal vector by requiring zero dot products with both spanning vectors.

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so the lines are parallel. 2. The point \((1, 0, 3)\) is not on \(h\): the first coordinate would require \(k=\frac{1}{2}\), while the second would require \(k=-1\). Thus the lines are distinct parallel lines. 3. Use \(\vec{u}=\begin{pmatrix}2\\-1\\4\end{pmatrix}\) and the vector between the given points, \(\vec{v}=\begin{pmatrix}2\\2\\-2\end{pmatrix}\), as spanning vectors. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(2a-b+4c=0\) and \(2a+2b-2c=0\). One solution is \(\vec{n}=\begin{pmatrix}-1\\2\\1\end{pmatrix}\). 4. Using \((1, 0, 3)\), the plane equation is \(F:-x+2y+z=2\).

a) The lines are distinct and parallel. b) One equation of their common plane is \(F:-x+2y+z=2\).

- Test whether one parameter value satisfies all three coordinates. - A line and a point outside it determine one plane. - Use the line direction and a vector from the line to the point as spanning vectors. - Find a normal vector by requiring zero dot products with both spanning vectors.

1. The first coordinate would require \(0=2+t\), so \(t=-2\). This value gives \(y=5-2\cdot(-2)=9\), not \(1\). Therefore, \(A\notin h\). A line and a point not on it determine a unique plane. 2. Use the line direction \(\vec{u}=\begin{pmatrix}1\\-2\\3\end{pmatrix}\) and the connecting vector \(\vec{v}=\begin{pmatrix}-2\\-4\\5\end{pmatrix}\). Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(a-2b+3c=0\) and \(-2a-4b+5c=0\). One solution is \(\vec{n}=\begin{pmatrix}2\\-11\\-8\end{pmatrix}\). 3. Using point \(A\), the plane equation is \(E:2x-11y-8z=-43\).

a) \(A\notin h\), so the line and the outside point determine a unique plane. b) \(E:2x-11y-8z=-43\)

- Two lines determine a plane when they intersect or are distinct and parallel. - First compare their direction vectors. - If they are not parallel, solve for a possible intersection point. - Use one point and both line directions as the plane's parametric data.

1. The direction vectors are not scalar multiples, so the lines are not parallel. 2. Equating coordinates gives \(r=1\) from the \(x\)-coordinates and \(s=1\) from the \(y\)-coordinates. The \(z\)-coordinates then both equal \(2\), so the lines intersect at \(S=(3, 1, 2)\). 3. Two intersecting, nonparallel lines determine a unique plane. One parametric equation is \(E:\vec{x}=\begin{pmatrix}2\\1\\3\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\-1\end{pmatrix}+\sigma\begin{pmatrix}0\\2\\1\end{pmatrix}\).

Yes. The lines intersect at \(S=(3, 1, 2)\), so they determine a unique plane. One equation is \(E:\vec{x}=\begin{pmatrix}2\\1\\3\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\-1\end{pmatrix}+\sigma\begin{pmatrix}0\\2\\1\end{pmatrix}\).

- Two distinct lines determine a plane when they intersect or are parallel. - First compare the direction vectors. - If the lines are not parallel, equate their coordinates to test for an intersection. - What does an inconsistent system imply about a common plane?

1. The direction vectors are not scalar multiples, so the lines are not parallel. 2. Equating coordinates gives \(2\lambda=2+\mu\), \(5-\lambda=1\), and \(2+\lambda=4+3\mu\). 3. The second equation gives \(\lambda=4\), and the first then gives \(\mu=6\). 4. These values do not satisfy the third equation because \(2+4=6\), while \(4+3\cdot6=22\). 5. Therefore, the lines are skew. Skew lines do not lie in one plane, so they do not determine a common plane.

No. The lines are skew, so they do not determine a common plane.

- Use one point and two connecting vectors for parametric form. - Find a normal vector by requiring zero dot products with both spanning vectors. - Expand the point-normal equation to obtain Cartesian form. - Substitute the test point into the Cartesian equation.

1. Two spanning vectors are \(\overrightarrow{PQ}=\begin{pmatrix}-2\\3\\-1\end{pmatrix}\) and \(\overrightarrow{PR}=\begin{pmatrix}1\\-1\\1\end{pmatrix}\). Thus \(E:\vec{x}=\begin{pmatrix}4\\0\\1\end{pmatrix}+r\begin{pmatrix}-2\\3\\-1\end{pmatrix}+s\begin{pmatrix}1\\-1\\1\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(-2a+3b-c=0\) and \(a-b+c=0\). One solution is \(\vec{n}=\begin{pmatrix}2\\1\\-1\end{pmatrix}\), so a point-normal equation is \(\left(\vec{x}-\begin{pmatrix}4\\0\\1\end{pmatrix}\right)\cdot\begin{pmatrix}2\\1\\-1\end{pmatrix}=0\). 3. Expanding gives \(E:2x+y-z=7\). 4. Substituting \(S\) gives \(2\cdot1+4-(-1)=7\), so \(S\in E\).

Parametric form: \(E:\vec{x}=\begin{pmatrix}4\\0\\1\end{pmatrix}+r\begin{pmatrix}-2\\3\\-1\end{pmatrix}+s\begin{pmatrix}1\\-1\\1\end{pmatrix}\) Point-normal form: \(\left(\vec{x}-\begin{pmatrix}4\\0\\1\end{pmatrix}\right)\cdot\begin{pmatrix}2\\1\\-1\end{pmatrix}=0\) Cartesian form: \(E:2x+y-z=7\) The point \(S=(1, 4, -1)\) lies in \(E\).

- Use the intersection point and both line directions for parametric form. - Find a normal vector by requiring zero dot products with both spanning vectors. - Derive Cartesian form from point-normal form. - Test the point in the Cartesian equation.

1. The common point and the two line directions give \(E:\vec{x}=\begin{pmatrix}1\\0\\2\end{pmatrix}+r\begin{pmatrix}2\\1\\0\end{pmatrix}+s\begin{pmatrix}0\\-1\\1\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both direction vectors. Then \(2a+b=0\) and \(-b+c=0\). One solution is \(\vec{n}=\begin{pmatrix}1\\-2\\-2\end{pmatrix}\). Thus a point-normal equation is \(\left(\vec{x}-\begin{pmatrix}1\\0\\2\end{pmatrix}\right)\cdot\begin{pmatrix}1\\-2\\-2\end{pmatrix}=0\). 3. The Cartesian equation is \(E:x-2y-2z=-3\). 4. For \(T\), \(5-2\cdot1-2\cdot3=-3\), so \(T\in E\).

Parametric form: \(E:\vec{x}=\begin{pmatrix}1\\0\\2\end{pmatrix}+r\begin{pmatrix}2\\1\\0\end{pmatrix}+s\begin{pmatrix}0\\-1\\1\end{pmatrix}\) Point-normal form: \(\left(\vec{x}-\begin{pmatrix}1\\0\\2\end{pmatrix}\right)\cdot\begin{pmatrix}1\\-2\\-2\end{pmatrix}=0\) Cartesian form: \(E:x-2y-2z=-3\) The point \(T=(5, 1, 3)\) lies in \(E\).

- Use the line direction and a vector from the line to the outside point as spanning vectors. - Confirm that the point is not already on the line. - Find a normal vector by requiring zero dot products with both spanning vectors. - A point-normal equation uses one point on the plane and a vector perpendicular to the plane.

1. Use the line direction \(\vec{u}=\begin{pmatrix}1\\-2\\2\end{pmatrix}\) and the vector from the line's given point to \(P\), \(\vec{v}=\begin{pmatrix}4\\-3\\0\end{pmatrix}\). They are linearly independent, so \(F:\vec{x}=\begin{pmatrix}0\\3\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}+\mu\begin{pmatrix}4\\-3\\0\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(a-2b+2c=0\) and \(4a-3b=0\). One solution is \(\vec{n}=\begin{pmatrix}6\\8\\5\end{pmatrix}\). Therefore, a point-normal equation is \(\left(\vec{x}-\begin{pmatrix}0\\3\\1\end{pmatrix}\right)\cdot\begin{pmatrix}6\\8\\5\end{pmatrix}=0\).

1. \(F:\vec{x}=\begin{pmatrix}0\\3\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}+\mu\begin{pmatrix}4\\-3\\0\end{pmatrix}\) 2. \(\left(\vec{x}-\begin{pmatrix}0\\3\\1\end{pmatrix}\right)\cdot\begin{pmatrix}6\\8\\5\end{pmatrix}=0\)

- Substitute a general point of the line into the plane equation. - What two conditions characterize skew lines? - Compare the direction vectors to rule out parallel lines. - Equate the line equations and look for a contradiction that does not depend on \(k\).

1. A point on \(g\) has coordinates \((\lambda,2,\lambda)\). Substitution into the plane equation gives \(\lambda+2-\lambda=2\), which is true for every \(\lambda\). Therefore, \(g\) lies in \(E\). 2. The direction vectors \(\begin{pmatrix}1\\0\\1\end{pmatrix}\) and \(\begin{pmatrix}0\\k\\1\end{pmatrix}\) cannot be scalar multiples because their first coordinates are \(1\) and \(0\). Thus the lines are never parallel. 3. Equating coordinates gives \(\lambda=1\), \(2=k\mu\), and \(\lambda=1+\mu\). 4. The first and third equations give \(\mu=0\). The second equation would then require \(2=0\), a contradiction for every \(k\). 5. The lines are never parallel and never intersect, so they are skew for every \(k\in\mathbb{R}\).

a) Substitution gives \(\lambda+2-\lambda=2\), so every point of \(g\) lies in \(E\). b) The lines are not parallel, and their intersection system leads to the contradiction \(2=0\). Therefore, they are skew for every \(k\in\mathbb{R}\).

- How can a known coordinate be used to find the line parameter? - What equation represents the condition that two coordinates are equal? - Once the parameter is known, substitute it into all three coordinate expressions.

1. A direction vector is \(\overrightarrow{AB}=\begin{pmatrix}2\\-3\\4\end{pmatrix}\), so \(h\) can be written as \(\vec{x}=\begin{pmatrix}1\\4\\-2\end{pmatrix}+r\begin{pmatrix}2\\-3\\4\end{pmatrix}\). 2. For \(T\), the \(y\)-coordinate gives \(10=4-3r\), so \(r=-2\). The \(z\)-coordinate checks because \(-2+4\cdot(-2)=-10\). Therefore, \(k=1+2\cdot(-2)=-3\). 3. Require \(x=z\): \(1+2r=-2+4r\). Thus \(3=2r\), so \(r=\frac{3}{2}\). Substitution gives \(\left(4,-\frac{1}{2},4\right)\).

a) \(h:\vec{x}=\begin{pmatrix}1\\4\\-2\end{pmatrix}+r\begin{pmatrix}2\\-3\\4\end{pmatrix}\) b) \(k=-3\) c) Yes. The point is \(\left(4,-\frac{1}{2},4\right)\).

- Write one equation for each coordinate. - A point lies in the plane only when one pair of parameter values satisfies all three equations. - For each point with a missing coordinate, first use the two known coordinates to find \(r\) and \(s\). - Then substitute the parameters into the remaining coordinate expression.

1. For \(P\), the coordinate equations are \(1+r+2s=4\), \(r-s=0\), and \(2+r=3\). The third equation gives \(r=1\), and the second gives \(s=1\). These values satisfy the first equation, so \(P\) lies in \(E\). 2. For \(Q\), the third equation gives \(r=3\), and the second gives \(s=2\). The first coordinate would then be \(1+3+2\cdot2=8\), not \(5\). Therefore, \(Q\) does not lie in \(E\). 3. For \(D\), the third coordinate gives \(r=2\). The first coordinate gives \(2=1+2+2s\), so \(s=-\frac{1}{2}\). Thus \(y=r-s=2-\left(-\frac{1}{2}\right)=\frac{5}{2}\). 4. For \(F\), the third coordinate gives \(r=4\). The second gives \(3=4-s\), so \(s=1\). Then \(x=1+4+2\cdot1=7\).

a) Point \(P\) lies in \(E\); point \(Q\) does not. b) \(y=\frac{5}{2}\) and \(x=7\)

- First compare the direction vectors. - Equate corresponding coordinates to form a system. - Solve two equations for the line parameters. - Verify the parameter values in the remaining equation before finding the point.

1. The direction vectors are not scalar multiples, so the lines are not parallel or identical. 2. Equating coordinates gives \(r=2s\), \(3-r=s\), and \(-1+2r=6-3s\). 3. Solving the first two equations gives \(s=1\) and \(r=2\). 4. These values satisfy the third equation because \(-1+2\cdot2=6-3\cdot1=3\). 5. Substituting \(r=2\) into \(g\) gives the intersection point \((7, 0, 3)\).

The lines intersect at \((7, 0, 3)\).

- Equate the line equations coordinate by coordinate. - Begin with the equations that do not contain \(a\). - Use those equations to find the line parameters. - Substitute the parameters into the remaining equation to determine \(a\).

1. Equating coordinates gives \(4+r=1+2s\), \(a+2r=1+s\), and \(3+r=2\). 2. The third equation gives \(r=-1\). 3. Substituting into the first equation gives \(3=1+2s\), so \(s=1\). 4. The second equation then gives \(a-2=2\), so \(a=4\). 5. Substituting \(s=1\) into \(h\) gives the intersection point \((3, 2, 2)\).

The lines intersect when \(a=4\), at \((3, 2, 2)\).

- What is special about the equation of a line through the origin? - The same parameter value must satisfy all three coordinate equations. - In part c, use the fixed \(z\)-coordinate to find the parameter first.

1. Because the line passes through the origin, write \(h:\vec{x}=s\begin{pmatrix}4\\-2\\6\end{pmatrix}\). 2. For \(P\), all three coordinates give \(s=-\frac{3}{2}\), so \(P\) lies on \(h\). For \(Q\), the \(x\)- and \(y\)-coordinates give \(s=\frac{5}{2}\), but the \(z\)-coordinate gives \(s=2\). Therefore, \(Q\) does not lie on \(h\). 3. For \(S\), \(6s=21\), so \(s=\frac{7}{2}\). Thus \(x=4\cdot\frac{7}{2}=14\) and \(y=-2\cdot\frac{7}{2}=-7\). 4. For \(R\), the \(z\)-coordinate forces \(s=3\). Then the line has \(x=12\) and \(y=-6\). The \(x\)-coordinate would require \(k=12\), while the \(y\)-coordinate \(k+6=-6\) would require \(k=-12\). These conditions contradict each other, so no such \(k\) exists.

a) Point \(P\) lies on \(h\); point \(Q\) does not. b) \(x=14\) and \(y=-7\) c) The \(z\)-coordinate forces \(s=3\), which requires both \(k=12\) and \(k=-12\). Therefore, no value of \(k\) works.

- How do the coordinates of a point on the line depend on the parameter? - Write the sum of the three coordinates in terms of the parameter. - Use the magnitude formula for a vector. - Can two points on opposite sides of the origin have the same distance from it?

1. Because the line passes through the origin, write \(h:\vec{x}=s\begin{pmatrix}3\\-2\\6\end{pmatrix}\). 2. A point on \(h\) has coordinates \((3s, -2s, 6s)\), whose sum is \(7s\). Solve \(7s=21\) to get \(s=3\). Thus \(B=(9, -6, 18)\). 3. The magnitude of the direction vector is \(\sqrt{3^2+(-2)^2+6^2}=7\). Therefore, a point with parameter \(s\) is \(7|s|\) units from the origin. Solve \(7|s|=35\), so \(s=5\) or \(s=-5\). The points are \((15, -10, 30)\) and \((-15, 10, -30)\).

a) \(h:\vec{x}=s\begin{pmatrix}3\\-2\\6\end{pmatrix}\) b) \(B=(9, -6, 18)\) c) \(C=(15, -10, 30)\) or \(C=(-15, 10, -30)\)

- When does a point satisfy a line given by a coordinate equation? - Write the parametric equation as separate equations for \(x\) and \(y\). - Compare the components of the direction vector with the coordinates of \(M\).

1. The point \((-b, a)\) satisfies the line equation because \(a(-b)+b(a)=-ab+ab=0\). Since \((-b, a)\ne(0, 0)\), the vector \(\begin{pmatrix}-b\\a\end{pmatrix}\) points along the line and is a valid direction vector. 2. The parametric equation gives \(x=5t\) and \(y=2t\). Eliminating \(t\) yields \(y=\frac{2}{5}x\), or equivalently \(2x-5y=0\). 3. Substitute \(M(k,6)\) into \(2x-5y=0\): \(2k-5\cdot6=0\). Therefore, \(2k=30\), so \(k=15\).

a) \(a(-b)+b(a)=0\), so \(\begin{pmatrix}-b\\a\end{pmatrix}\) is a direction vector. b) \(2x-5y=0\) c) \(k=15\)

- How can you find a direction vector from two points? - Which coordinate represents altitude? - How can the known altitude be used to find the parameter? - Once the parameter is known, how do you find the other coordinates?

1. A direction vector is \(\overrightarrow{AB}=\begin{pmatrix}30\\30\\-30\end{pmatrix}\). Therefore, \(h\) can be written as \(\vec{x}=\begin{pmatrix}10\\20\\100\end{pmatrix}+t\begin{pmatrix}30\\30\\-30\end{pmatrix}\). 2. Set the altitude equal to \(85\): \(100-30t=85\). Then \(t=\frac{1}{2}\). 3. Substitute \(t=\frac{1}{2}\): \(x=10+30\cdot\frac{1}{2}=25\) and \(y=20+30\cdot\frac{1}{2}=35\). The point is \((25, 35, 85)\).

a) \(h:\vec{x}=\begin{pmatrix}10\\20\\100\end{pmatrix}+t\begin{pmatrix}30\\30\\-30\end{pmatrix}\) b) \((25, 35, 85)\)

- What must be true about direction vectors of parallel lines? - Which position vector is used for a line through the origin? - How do you translate a point by a given vector? - How can you verify that a point lies on a parametric line?

1. A direction vector for \(g\) is \(\overrightarrow{AB}=\begin{pmatrix}3\\2\\2\end{pmatrix}\). Since \(h\) is parallel to \(g\) and passes through the origin, \(h:\vec{x}=s\begin{pmatrix}3\\2\\2\end{pmatrix}\). 2. Translating \(P\) by \(\overrightarrow{OA}=\begin{pmatrix}2\\-1\\5\end{pmatrix}\) gives \(\overrightarrow{OP'}=\begin{pmatrix}6\\4\\4\end{pmatrix}+\begin{pmatrix}2\\-1\\5\end{pmatrix}=\begin{pmatrix}8\\3\\9\end{pmatrix}\). Thus \(P'=(8, 3, 9)\). 3. Line \(g\) can be written as \(\vec{x}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+r\begin{pmatrix}3\\2\\2\end{pmatrix}\). For \(r=2\), this gives \(\begin{pmatrix}8\\3\\9\end{pmatrix}\), so \(P'\) lies on \(g\).

a) \(h:\vec{x}=s\begin{pmatrix}3\\2\\2\end{pmatrix}\) b) \(P'=(8, 3, 9)\). Substituting \(r=2\) into \(g:\vec{x}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+r\begin{pmatrix}3\\2\\2\end{pmatrix}\) gives \(P'\).

- Compare both the position points and direction vectors. - Substitute \(Q\) into each equation and solve for its parameter. - For the same point, set the two parameter displacements equal. - How does multiplying a direction vector by a negative number affect orientation?

1. The equations use the same position point, and the direction vector of \(h\) is \(-2\) times the direction vector of \(g\). Therefore, they describe the same line. 2. In equation \(g\), the first coordinate gives \(3+r=5\), so \(r=2\). The other coordinates also produce \((5, 5, 2)\). 3. In equation \(h\), the first coordinate gives \(3-2s=5\), so \(s=-1\). The other coordinates also produce \((5, 5, 2)\). 4. For the same point, \(r\vec{v}_g=s\vec{v}_h=s(-2\vec{v}_g)\). Thus \(r=-2s\), or \(s=-\frac{1}{2}r\). The negative sign reverses direction, and the factor \(2\) rescales the parameter.

a) The equations describe the same line because they have the same position point and \(\vec{v}_h=-2\vec{v}_g\). b) \(r=2\) and \(s=-1\) c) \(r=-2s\), or equivalently \(s=-\frac{1}{2}r\).

- First find a direction vector from \(A\) to \(B\). - A matching equation needs a position point on \(g\). - Its direction vector must also be a nonzero scalar multiple of the direction of \(g\). - Check both conditions for every equation.

1. A direction vector for \(g\) is \(\overrightarrow{AB}=\begin{pmatrix}3\\3\\-3\end{pmatrix}\), so any valid direction vector must be a nonzero scalar multiple of \(\begin{pmatrix}1\\1\\-1\end{pmatrix}\). 2. For \(h\), the position point is \(B\), and its direction vector is \(-\frac{1}{3}\overrightarrow{AB}\). Therefore, \(h\) is the same line. 3. For \(k\), the position point \((8, 5, -3)\) lies on \(g\), but \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) is not parallel to \(\overrightarrow{AB}\). Therefore, \(k\) is a different line. 4. For \(m\), the position point \((11, 8, -6)\) lies on \(g\), and its direction vector is \(\frac{1}{2}\begin{pmatrix}3\\3\\-3\end{pmatrix}\). Therefore, \(m\) is the same line.

a) same line b) different line c) same line

- How can you find another point on line \(g\)? - First find the magnitude of the original direction vector. - What scalar both reverses the vector and changes its magnitude to the target value? - A valid parametric equation needs both a point on the line and a parallel direction vector.

1. Choose \(t=1\) in the original equation. This gives a new point on the line: \(\begin{pmatrix}4\\-2\\5\end{pmatrix}+\begin{pmatrix}2\\6\\-4\end{pmatrix}=\begin{pmatrix}6\\4\\1\end{pmatrix}\). 2. The original direction vector has magnitude \(\sqrt{2^2+6^2+(-4)^2}=\sqrt{56}=2\sqrt{14}\). 3. To obtain magnitude \(3\sqrt{14}\) and reverse the direction, multiply by \(-\frac{3}{2}\): \(-\frac{3}{2}\begin{pmatrix}2\\6\\-4\end{pmatrix}=\begin{pmatrix}-3\\-9\\6\end{pmatrix}\). 4. Therefore, one valid equation is \(\vec{x}=\begin{pmatrix}6\\4\\1\end{pmatrix}+s\begin{pmatrix}-3\\-9\\6\end{pmatrix}\).

One possible equation is \(\vec{x}=\begin{pmatrix}6\\4\\1\end{pmatrix}+s\begin{pmatrix}-3\\-9\\6\end{pmatrix}\), where \(s\in\mathbb{R}\).

- How do you find a point when the parameter is known? - What coordinate condition identifies the \(xz\)-plane? - How can a known coordinate be used to find the parameter? - Write the vector equation as separate equations for \(x\), \(y\), and \(z\).

1. For \(r=4\), \(A=\begin{pmatrix}4\\-2\\1\end{pmatrix}+4\begin{pmatrix}-2\\3\\2\end{pmatrix}=\begin{pmatrix}-4\\10\\9\end{pmatrix}\). 2. Set the \(x\)-coordinate equal to \(10\): \(4-2r=10\), so \(r=-3\). Then \(y=-2+3\cdot(-3)=-11\) and \(z=1+2\cdot(-3)=-5\). Thus \(B=(10, -11, -5)\). 3. A point in the \(xz\)-plane has \(y=0\). Solve \(-2+3r=0\), so \(r=\frac{2}{3}\). Then \(x=4-2\cdot\frac{2}{3}=\frac{8}{3}\) and \(z=1+2\cdot\frac{2}{3}=\frac{7}{3}\). Therefore, \(S=\left(\frac{8}{3},0,\frac{7}{3}\right)\).

a) \(A=(-4, 10, 9)\) b) \(r=-3\) and \(B=(10, -11, -5)\) c) \(r=\frac{2}{3}\) and \(S=\left(\frac{8}{3},0,\frac{7}{3}\right)\)

- A position vector must represent a point that lies on the line. - Substitute each point and check whether one parameter value works in all coordinates. - How does multiplying a direction vector by a scalar affect its direction and magnitude? - A line has infinitely many equivalent parametric equations.

1. For \(P\), the \(x\)-coordinate gives \(t=2\), and the other coordinates check: \(4-2=2\) and \(-2+3\cdot2=4\). Thus \(P\) lies on \(g\). 2. For \(Q\), the \(x\)-coordinate gives \(t=1\), and the other coordinates check: \(4-1=3\) and \(-2+3=1\). Thus \(Q\) lies on \(g\). 3. For \(R\), the \(x\)-coordinate gives \(t=-1\). Then the \(y\)-coordinate is \(5\), but the \(z\)-coordinate would be \(-5\ne-4\). Thus \(R\) does not lie on \(g\). 4. Multiplying the original direction vector by \(-3\) reverses its orientation and triples its magnitude: \(-3\begin{pmatrix}2\\-1\\3\end{pmatrix}=\begin{pmatrix}-6\\3\\-9\end{pmatrix}\). 5. Therefore, one new equation is \(g:\vec{x}=\begin{pmatrix}3\\3\\1\end{pmatrix}+s\begin{pmatrix}-6\\3\\-9\end{pmatrix}\).

a) Points \(P\) and \(Q\) lie on \(g\); point \(R\) does not. b) \(g:\vec{x}=\begin{pmatrix}3\\3\\1\end{pmatrix}+s\begin{pmatrix}-6\\3\\-9\end{pmatrix}\)

- Multiplying a direction vector by any nonzero scalar does not change the line. - Any point on the line can serve as the position vector. - How do you find the midpoint of two points? - What sign must the scalar have to reverse a vector's orientation?

1. The vector \(\overrightarrow{PQ}=\begin{pmatrix}6\\-4\end{pmatrix}\). 2. Using \(Q\) as the position vector gives \(\vec{x}=\begin{pmatrix}4\\1\end{pmatrix}+t\begin{pmatrix}6\\-4\end{pmatrix}\). 3. A vector with opposite orientation and twice the magnitude is \(-2\overrightarrow{PQ}=\begin{pmatrix}-12\\8\end{pmatrix}\). Using \(P\) as the position vector gives \(\vec{x}=\begin{pmatrix}-2\\5\end{pmatrix}+t\begin{pmatrix}-12\\8\end{pmatrix}\). 4. The midpoint is \(M=\left(\frac{-2+4}{2}, \frac{5+1}{2}\right)=(1, 3)\). Using \(M\) as the position vector gives \(\vec{x}=\begin{pmatrix}1\\3\end{pmatrix}+t\begin{pmatrix}6\\-4\end{pmatrix}\).

1. \(\vec{x}=\begin{pmatrix}4\\1\end{pmatrix}+t\begin{pmatrix}6\\-4\end{pmatrix}\) 2. \(\vec{x}=\begin{pmatrix}-2\\5\end{pmatrix}+t\begin{pmatrix}-12\\8\end{pmatrix}\) 3. \(\vec{x}=\begin{pmatrix}1\\3\end{pmatrix}+t\begin{pmatrix}6\\-4\end{pmatrix}\)

- Two equations describe the same line only when both their position and direction conditions agree. - Find \(\overrightarrow{AB}\). - Does each proposed position point lie on the line through \(A\) and \(B\)? - Is each proposed direction vector a nonzero scalar multiple of \(\overrightarrow{AB}\)? - Check whether \((2, 2, 0)\) has a special relationship to \(A\) and \(B\).

1. A direction vector from \(A\) to \(B\) is \(\overrightarrow{AB}=\begin{pmatrix}-2\\4\\4\end{pmatrix}\). 2. Mia uses point \(A\), which lies on the line. Her direction vector is \(\frac{1}{2}\overrightarrow{AB}\), so her equation is correct. 3. Tom uses \((2, 2, 0)\), the midpoint of \(A\) and \(B\), so his position point lies on the line. His direction vector is \(-\frac{1}{2}\overrightarrow{AB}\), so his equation is also correct.

Both equations are correct. Mia uses point \(A\) and a direction vector equal to \(\frac{1}{2}\overrightarrow{AB}\). Tom uses the midpoint of \(A\) and \(B\) and a direction vector equal to \(-\frac{1}{2}\overrightarrow{AB}\).

- How do you find the point where a line crosses the \(y\)-axis? - What normal vector is given by the coefficients in \(ax+by=c\)? - How can you find a vector perpendicular to a given vector? - Use the magnitude formula to check the required length.

1. At the \(y\)-intercept, \(x=0\). Then \(2y=6\), so \(y=3\). Thus \(\vec{p}=\begin{pmatrix}0\\3\end{pmatrix}\). 2. A normal vector to the line is \(\begin{pmatrix}3\\2\end{pmatrix}\). A perpendicular direction vector is \(\begin{pmatrix}2\\-3\end{pmatrix}\), since \(3\cdot2+2\cdot(-3)=0\). 3. Its magnitude is \(\sqrt{2^2+(-3)^2}=\sqrt{13}\), and its \(x\)-component is positive. Therefore, \(\vec{x}=\begin{pmatrix}0\\3\end{pmatrix}+t\begin{pmatrix}2\\-3\end{pmatrix}\).

\(\vec{x}=\begin{pmatrix}0\\3\end{pmatrix}+t\begin{pmatrix}2\\-3\end{pmatrix}\), where \(t\in\mathbb{R}\)

- What coordinate is zero at each axis intercept? - How are direction vectors of perpendicular lines related in the plane? - Write the parametric equation as separate equations for \(x\) and \(y\).

1. For the \(x\)-intercept, set \(y=0\): \(4-3t=0\), so \(t=\frac{4}{3}\). Then \(x=1+2\cdot\frac{4}{3}=\frac{11}{3}\), giving \(S_1=\left(\frac{11}{3},0\right)\). 2. For the \(y\)-intercept, set \(x=0\): \(1+2t=0\), so \(t=-\frac{1}{2}\). Then \(y=4-3\cdot\left(-\frac{1}{2}\right)=\frac{11}{2}\), giving \(S_2=\left(0,\frac{11}{2}\right)\). 3. Substitute \(a=\frac{11}{3}\) and \(b=\frac{11}{2}\): \(\frac{x}{11/3}+\frac{y}{11/2}=1\), equivalently \(\frac{3x}{11}+\frac{2y}{11}=1\). 4. A direction vector perpendicular to \(\begin{pmatrix}2\\-3\end{pmatrix}\) is \(\begin{pmatrix}3\\2\end{pmatrix}\), since \(2\cdot3+(-3)\cdot2=0\). Because \(h\) passes through the origin, \(h:\vec{x}=s\begin{pmatrix}3\\2\end{pmatrix}\).

1. \(S_1=\left(\frac{11}{3},0\right)\) and \(S_2=\left(0,\frac{11}{2}\right)\) 2. \(\frac{x}{11/3}+\frac{y}{11/2}=1\), equivalently \(\frac{3x}{11}+\frac{2y}{11}=1\) 3. \(h:\vec{x}=s\begin{pmatrix}3\\2\end{pmatrix}\)

- How are a line's direction vector and normal vector related? - Can you separate each vector equation into equations for \(x\) and \(y\)? - What happens when you eliminate the parameter? - In part 2, separate the constant vector from the parameter-dependent vector.

1. For \(g_1\), a normal vector perpendicular to \(\begin{pmatrix}3\\5\end{pmatrix}\) is \(\begin{pmatrix}5\\-3\end{pmatrix}\). Using point \((-1, 4)\), \(5\cdot(-1)-3\cdot4=-17\). Therefore, \(5x-3y=-17\). 2. Rewrite \(g_2\) as \(\vec{x}=\begin{pmatrix}3\\0\end{pmatrix}+s\begin{pmatrix}2\\-1\end{pmatrix}\). A normal vector is \(\begin{pmatrix}1\\2\end{pmatrix}\). Using point \((3, 0)\) gives \(x+2y=3\). 3. For \(g_3\), the direction vector is horizontal, so every point has \(y=2\). Thus the coordinate equation is \(y=2\).

1. \(g_1:5x-3y=-17\) 2. \(g_2:x+2y=3\) 3. \(g_3:y=2\)

- How can you find a direction vector from two points? - What is the relationship between a direction vector and a normal vector? - What does parallelism imply about the coefficients in a coordinate equation? - What must \(c\) be in \(ax+by=c\) for the line to pass through the origin?

1. A direction vector is \(\overrightarrow{AB}=\begin{pmatrix}-2\\4\end{pmatrix}\). Therefore, \(g:\vec{x}=\begin{pmatrix}3\\-2\end{pmatrix}+t\begin{pmatrix}-2\\4\end{pmatrix}\). 2. A normal vector perpendicular to \(\begin{pmatrix}-2\\4\end{pmatrix}\) is \(\begin{pmatrix}2\\1\end{pmatrix}\). Thus the equation has the form \(2x+y=c\). Substituting \(A(3, -2)\) gives \(c=2\cdot3-2=4\), so \(g:2x+y=4\). 3. A parallel line has the same normal vector, so \(h\) has the form \(2x+y=c\). Since it passes through \((0, 0)\), \(c=0\). Therefore, \(h:2x+y=0\).

1. \(g:\vec{x}=\begin{pmatrix}3\\-2\end{pmatrix}+t\begin{pmatrix}-2\\4\end{pmatrix}\) 2. \(g:2x+y=4\) 3. \(h:2x+y=0\)

- What condition must a point satisfy to lie on a parametric line? - How can a known coordinate be used to find the parameter? - Must the same parameter value work in all three coordinate equations? - What conclusion follows if two known coordinates require different parameter values?

1. For \(A\), the \(y\)-coordinate gives \(-2+r=0\), so \(r=2\). Then \(x=3+2\cdot2=7\) and \(z=1-3\cdot2=-5\). Thus \(A=(7, 0, -5)\) lies on \(g\). 2. For \(B\), the \(x\)-coordinate gives \(3+2r=7\), so \(r=2\). The \(z\)-coordinate would then be \(1-3\cdot2=-5\), not \(-4\). Equivalently, \(z=-4\) would require \(r=\frac{5}{3}\). Since the known coordinates require different parameter values, no value of \(y\) can place \(B\) on \(g\).

a) Yes. \(x=7\) and \(z=-5\), so \(A=(7, 0, -5)\). b) No. The known coordinates require \(r=2\) and \(r=\frac{5}{3}\), respectively.

- First check whether the direction vectors are scalar multiples. - If the directions are parallel, what two relationships remain possible? - Test one line's position point in the other line. - A shared point and parallel directions imply what relationship?

1. The direction vector of \(h\) is \(-1.5\) times the direction vector of \(g\), so the lines are parallel or identical. 2. Test the position point \((2, 6, -17)\) from \(h\) in line \(g\). The first coordinate gives \(2=5+2\lambda\), so \(\lambda=-1.5\). 3. This value also gives \(0-4\cdot(-1.5)=6\) and \(-2+10\cdot(-1.5)=-17\). Thus the point lies on \(g\). 4. Since the lines have parallel directions and share a point, they are identical.

The lines are identical.

- Parallel direction vectors are scalar multiples; use coordinates without \(k\) to find the scale factor first. - After finding \(k\), test whether one line's position point lies on the other line. - Checking only one coordinate is not enough for a point test.

1. Parallel direction vectors must be scalar multiples. From \(-9=3m\), the scale factor is \(m=-3\). The third coordinates are consistent because \(6=(-3)\cdot(-2)\). 2. The second coordinates give \(12=-3k\), so \(k=-4\). 3. For \(k=-4\), test the point \((1, 2, 0)\) from \(g_k\) in \(h\). The first coordinate gives \(1=4-9s\), so \(s=\frac{1}{3}\). 4. This value gives a second coordinate of \(-1+12\cdot\left(\frac{1}{3}\right)=3\), not \(2\). The point is not on \(h\), so the parallel lines are distinct.

a) \(k=-4\) b) For \(k=-4\), the lines are distinct and parallel.

- Begin by substituting point \(R\) for \(\vec{x}\). - Which coordinate equation contains only one unknown? - Find the line parameter first. - Then use that value to solve the remaining equations in sequence.

1. Equating coordinates gives \(1=s+2\lambda\), \(5=r+\lambda s\), and \(-2=4-3\lambda\). 2. The third equation gives \(\lambda=2\). 3. Substitute \(\lambda=2\) into the first equation: \(1=s+4\), so \(s=-3\). 4. Substitute \(\lambda=2\) and \(s=-3\) into the second equation: \(5=r+2\cdot(-3)\), so \(r=11\).

\(r=11\) and \(s=-3\)

- Compare the direction vectors for each pair first. - For parallel directions, use a point test to distinguish identical lines from distinct parallel lines. - For nonparallel directions, equate coordinates and solve for the parameters. - Verify any parameter values in every coordinate equation. - If nonparallel lines have no common point, they are skew.

1. For \(g_1\) and \(g_2\), the direction vectors are not scalar multiples. Equating coordinates and solving the first two equations gives \(r=2\) and \(s=0\). These values satisfy the third equation, so the lines intersect at \((5, 4, 1)\). 2. For \(g_1\) and \(g_3\), the direction vector of \(g_3\) is \(-2\) times the direction vector of \(g_1\). Testing \((-3, 0, 0)\) on \(g_1\) gives inconsistent parameter values, so the lines are distinct and parallel. 3. For \(g_2\) and \(g_3\), the direction vectors are not scalar multiples. The first two coordinate equations give \(s=0\) and \(t=-2\). These values give \(z=1\) on \(g_2\) but \(z=-4\) on \(g_3\), so the remaining coordinate equation fails. Therefore, the lines are skew.

\(g_1\) and \(g_2\) intersect at \((5, 4, 1)\). \(g_1\) and \(g_3\) are distinct and parallel. \(g_2\) and \(g_3\) are skew.

- Use the coordinate equations that do not contain \(a\) to find the line parameters. - Substitute those parameter values into the remaining equation. - Parallelism depends on the direction vectors, not the position point. - Which relationship remains for nonparallel lines that do not intersect?

1. Equating coordinates gives \(4+\lambda=2+3\mu\), \(2\lambda=a+2\mu\), and \(3-\lambda=1+\mu\). 2. Solving the first and third equations gives \(\lambda=1\) and \(\mu=1\). 3. The second equation then gives \(2=a+2\), so \(a=0\). Substituting \(\lambda=1\) into \(g\) gives the intersection point \((5, 2, 2)\). 4. The direction vectors \(\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(\begin{pmatrix}3\\2\\1\end{pmatrix}\) are not scalar multiples, regardless of \(a\), so the lines are never parallel. 5. For \(a\neq0\), the lines are nonparallel and do not intersect, so they are skew.

a) The lines intersect when \(a=0\), at \((5, 2, 2)\). b) Their direction vectors are not scalar multiples. c) For \(a\neq0\), the lines are skew.

- Compare the direction vectors for each pair. - For parallel directions, test whether a position point lies on the other line. - For nonparallel directions, equate coordinates and solve for the parameters. - Check all coordinate equations before concluding that an intersection exists.

1. For \(g\) and \(h\), the direction vectors are not scalar multiples. Equating coordinates gives \(r=2\), \(s=0\), and the third equation is satisfied. Thus they intersect at \((3, 2, 5)\). 2. For \(g\) and \(k\), the direction vectors are scalar multiples. The point \((2, 0, 1)\) from \(k\) is not on \(g\), since the second coordinate would require \(2=0\). Therefore, the lines are distinct and parallel. 3. For \(h\) and \(k\), the direction vectors are not scalar multiples. The first two coordinate equations give \(t=-\frac{1}{2}\) and \(s=-2\), but these values do not satisfy the third equation. Therefore, the lines are skew.

\(g\) and \(h\) intersect at \((3, 2, 5)\). \(g\) and \(k\) are distinct and parallel. \(h\) and \(k\) are skew.

- First compare all direction vectors. - What does a system with infinitely many solutions mean for two parallel-direction lines? - What does an inconsistent system mean when the directions are parallel? - Every coordinate equation must be satisfied by a common point.

1. All three direction vectors are scalar multiples, so every pair is either identical or distinct and parallel. 2. Equating \(g\) and \(h\) produces three equivalent equations, each reducing to \(2r+s=-2\). The system has infinitely many solutions, so \(g\) and \(h\) are identical. 3. Equating \(g\) and \(k\) gives \(2r-3t=-1\) from the first two coordinates, but \(2r-3t=\frac{1}{3}\) from the third. The system is inconsistent, so \(g\) and \(k\) are distinct and parallel. 4. Therefore, \(g\) and \(h\) describe the same line, while \(k\) is a distinct parallel line.

Lines \(g\) and \(h\) are identical. Line \(k\) is distinct and parallel to both.

- Compare the direction vectors first. - If equating two lines produces the same equation twice, how many common points are there? - What does infinitely many common points imply about two lines? - Once \(g=h\) and \(h=k\), what follows about \(g\) and \(k\)?

1. The direction vectors are all scalar multiples, so the lines are parallel or identical. 2. Equating \(g\) and \(h\) gives \(2\lambda+\mu=1\) from both coordinates. The system has infinitely many solutions, so \(g\) and \(h\) are identical. 3. Equating \(h\) and \(k\) gives \(\mu-3\sigma=2\) from both coordinates. This system also has infinitely many solutions, so \(h\) and \(k\) are identical. 4. Therefore, all three equations describe the same line.

All three parametric equations describe the same line.

- Identical lines need parallel direction vectors and a shared point. - First compare the direction vectors. - Require the position point of \(h_{a,b}\) to lie on \(g\). - What relationships are possible when two lines have parallel directions?

1. The direction vector of \(h_{a,b}\) is \(-2\) times the direction vector of \(g\), so the lines are always parallel or identical. 2. For identical lines, the position point \((a,10,b)\) must lie on \(g\). The second coordinate gives \(10=1+3\lambda\), so \(\lambda=3\). 3. The first and third coordinates then give \(a=2+3=5\) and \(b=-3-2\cdot3=-9\). 4. Because the direction vectors are parallel for all \(a\) and \(b\), the lines can only be identical or distinct and parallel. Neither case has exactly one intersection point.

1. \(a=5\) and \(b=-9\) 2. The lines always have parallel direction vectors, so they can only be identical or distinct and parallel, never intersect at exactly one point.

- Determine when the direction vectors are scalar multiples. - For the parallel case, test whether a position point lies on the other line. - For the nonparallel case, compare the constant third coordinates. - Nonparallel lines with no common point have what relationship?

1. For parallel direction vectors, \(\begin{pmatrix}k+1\\1\\0\end{pmatrix}\) must equal \(c\begin{pmatrix}2\\1\\0\end{pmatrix}\). The second coordinate gives \(c=1\), so \(k+1=2\) and \(k=1\). 2. When \(k=1\), the lines have the same direction vector. The point \((3, 1, 1)\) on \(f_1\) lies on \(g\) when \(r=1\), so the lines are identical. 3. When \(k\neq1\), the direction vectors are not parallel. Any intersection would require equality of the third coordinates, \(1=k\), which is impossible. Therefore, the lines are skew.

For \(k=1\), the lines are identical. For \(k\neq1\), the lines are skew.

- Equate the line equations coordinate by coordinate. - Start with the equations that do not contain \(a\). - Use the resulting line parameters in the remaining equation. - Substitute one parameter into its line to find the common point.

1. Equating coordinates gives \(3+r=1+3s\), \(1-2r=7+as\), and \(4+2r=-2-2s\). 2. From the first equation, \(r=3s-2\). Substitution into the third equation gives \(s=-\frac{1}{4}\), and then \(r=-\frac{11}{4}\). 3. Substituting these values into the second equation gives \(\frac{13}{2}=7-\frac{a}{4}\), so \(a=2\). 4. Substituting \(r=-\frac{11}{4}\) into \(g\) gives the intersection point \(\left(\frac{1}{4},\frac{13}{2},-\frac{3}{2}\right)\).

a) \(a=2\) b) \(\left(\frac{1}{4}, \frac{13}{2}, -\frac{3}{2}\right)\)

- Compare the position points before doing any algebra. - Compare the direction vectors for scalar multiples. - Look for coordinates that stay constant along each line. - Nonparallel lines without a common point are skew.

1. In part a, the lines have the same position point and nonparallel direction vectors. Therefore, they intersect at their shared position point \((5, 2, -1)\). 2. In part b, every point on \(g\) has \(x=1\), while every point on \(h\) has \(x=2\), so the lines cannot intersect. Their direction vectors are not scalar multiples, so they are skew.

a) The lines intersect at \((5, 2, -1)\). b) The lines are skew.

- Compare the direction vectors first. - For parallel directions, test whether one position point lies on the other line. - A shared point makes parallel-direction lines identical. - The vector between position points can reveal whether the lines are the same.

1. In part a, the direction vectors are scalar multiples. The position point \((1, 1, 0)\) of \(g\) lies on \(h\) when \(\mu=1\). Therefore, the lines are identical. 2. In part b, the lines have the same direction vector, so they are parallel or identical. Their position-point difference is \(\begin{pmatrix}0\\0\\1\end{pmatrix}\), which is not a scalar multiple of \(\begin{pmatrix}1\\1\\1\end{pmatrix}\). Therefore, they are distinct and parallel.

a) The lines are identical. b) The lines are distinct and parallel.

- Which two points determine each requested line? - How do you find the midpoint of a segment? - What formula gives the centroid of a triangle in three dimensions? - Use one point as the position vector and subtract coordinates to get a direction vector.

1. The midpoint of \(\overline{AB}\) is \(M_{AB}=\left(\frac{6+0}{2}, \frac{0+6}{2}, 0\right)=(3, 3, 0)\). Since \(C\) is the origin, \(g:\vec{x}=r\begin{pmatrix}3\\3\\0\end{pmatrix}\). 2. The centroid of triangle \(ABC\) is \(S_{ABC}=\left(\frac{6+0+0}{3}, \frac{0+6+0}{3}, 0\right)=(2, 2, 0)\). 3. A direction vector from \(D\) to \(S_{ABC}\) is \(\begin{pmatrix}2\\2\\0\end{pmatrix}-\begin{pmatrix}3\\3\\9\end{pmatrix}=\begin{pmatrix}-1\\-1\\-9\end{pmatrix}\). Thus \(h:\vec{x}=\begin{pmatrix}3\\3\\9\end{pmatrix}+s\begin{pmatrix}-1\\-1\\-9\end{pmatrix}\).

a) \(g:\vec{x}=r\begin{pmatrix}3\\3\\0\end{pmatrix}\) b) \(h:\vec{x}=\begin{pmatrix}3\\3\\9\end{pmatrix}+s\begin{pmatrix}-1\\-1\\-9\end{pmatrix}\)

- Identify two points on each requested line. - Use a different parameter for each line. - The midpoint of a segment is found by averaging corresponding coordinates.

1. A direction vector from \(A\) to \(S\) is \(\begin{pmatrix}3\\3\\10\end{pmatrix}\), so \(g_{AS}:\vec{x}=\begin{pmatrix}2\\2\\0\end{pmatrix}+t\begin{pmatrix}3\\3\\10\end{pmatrix}\). 2. A direction vector from \(C\) to \(S\) is \(\begin{pmatrix}-3\\-3\\10\end{pmatrix}\), so \(g_{CS}:\vec{x}=\begin{pmatrix}8\\8\\0\end{pmatrix}+u\begin{pmatrix}-3\\-3\\10\end{pmatrix}\). 3. The midpoint of \(\overline{BC}\) is \(M_{BC}=\left(8, \frac{2+8}{2}, 0\right)=(8, 5, 0)\). A direction vector from \(S\) to \(M_{BC}\) is \(\begin{pmatrix}3\\0\\-10\end{pmatrix}\). Thus \(m:\vec{x}=\begin{pmatrix}5\\5\\10\end{pmatrix}+v\begin{pmatrix}3\\0\\-10\end{pmatrix}\).

a) \(g_{AS}:\vec{x}=\begin{pmatrix}2\\2\\0\end{pmatrix}+t\begin{pmatrix}3\\3\\10\end{pmatrix}\) \(g_{CS}:\vec{x}=\begin{pmatrix}8\\8\\0\end{pmatrix}+u\begin{pmatrix}-3\\-3\\10\end{pmatrix}\) b) \(m:\vec{x}=\begin{pmatrix}5\\5\\10\end{pmatrix}+v\begin{pmatrix}3\\0\\-10\end{pmatrix}\)

- How do you form a direction vector from two points? - How do you find the midpoint of a segment? - What formula gives the centroid from the three vertices? - What must be true about direction vectors of parallel lines?

1. A direction vector from \(B\) to \(C\) is \(\begin{pmatrix}-3\\-2\\4\end{pmatrix}\). Thus \(g_{BC}:\vec{x}=\begin{pmatrix}0\\5\\2\end{pmatrix}+r\begin{pmatrix}-3\\-2\\4\end{pmatrix}\). 2. The midpoint is \(M_{BC}=\left(-\frac{3}{2}, 4, 4\right)\). A direction vector from \(A\) to \(M_{BC}\) is \(\begin{pmatrix}-\frac{9}{2}\\3\\0\end{pmatrix}\), which is parallel to \(\begin{pmatrix}-3\\2\\0\end{pmatrix}\). Therefore, \(s_a:\vec{x}=\begin{pmatrix}3\\1\\4\end{pmatrix}+s\begin{pmatrix}-3\\2\\0\end{pmatrix}\). 3. The centroid is \(S=\left(\frac{3+0-3}{3}, \frac{1+5+3}{3}, \frac{4+2+6}{3}\right)=(0, 3, 4)\). 4. The midpoint of \(\overline{AC}\) is \(M_{AC}=(0, 2, 5)\). A direction vector for median \(s_b\) is \(\overrightarrow{BM_{AC}}=\begin{pmatrix}0\\-3\\3\end{pmatrix}\), which is parallel to \(\begin{pmatrix}0\\-1\\1\end{pmatrix}\). Using \(S\) as the position vector gives \(h:\vec{x}=\begin{pmatrix}0\\3\\4\end{pmatrix}+t\begin{pmatrix}0\\-1\\1\end{pmatrix}\).

a) \(g_{BC}:\vec{x}=\begin{pmatrix}0\\5\\2\end{pmatrix}+r\begin{pmatrix}-3\\-2\\4\end{pmatrix}\) b) \(M_{BC}=\left(-\frac{3}{2}, 4, 4\right)\) and \(s_a:\vec{x}=\begin{pmatrix}3\\1\\4\end{pmatrix}+s\begin{pmatrix}-3\\2\\0\end{pmatrix}\) c) \(S=(0, 3, 4)\) d) \(h:\vec{x}=\begin{pmatrix}0\\3\\4\end{pmatrix}+t\begin{pmatrix}0\\-1\\1\end{pmatrix}\)

- Use a vector relationship for opposite sides of a parallelogram. - Where do the diagonals of a parallelogram intersect? - What is the general form of a parametric line equation? - A direction vector may be simplified by a nonzero scalar factor.

1. In a parallelogram, \(\overrightarrow{AD}=\overrightarrow{BC}\). Thus \(D=A+(C-B)=(2, -1, 0)+(-2, 4, -2)=(0, 3, -2)\). 2. A direction vector from \(B\) to \(D\) is \(\begin{pmatrix}-6\\2\\-6\end{pmatrix}\), which is parallel to \(\begin{pmatrix}-3\\1\\-3\end{pmatrix}\). Therefore, \(g_{BD}:\vec{x}=\begin{pmatrix}6\\1\\4\end{pmatrix}+r\begin{pmatrix}-3\\1\\-3\end{pmatrix}\). 3. The diagonals bisect each other, so \(M\) is the midpoint of \(\overline{AC}\): \(M=(3, 2, 1)\). 4. The midpoint of \(\overline{CD}\) is \(M_{CD}=(2, 4, 0)\). A direction vector from \(M\) to \(M_{CD}\) is \(\begin{pmatrix}-1\\2\\-1\end{pmatrix}\). Thus \(k:\vec{x}=\begin{pmatrix}3\\2\\1\end{pmatrix}+t\begin{pmatrix}-1\\2\\-1\end{pmatrix}\).

a) \(D=(0, 3, -2)\) b) \(g_{BD}:\vec{x}=\begin{pmatrix}6\\1\\4\end{pmatrix}+r\begin{pmatrix}-3\\1\\-3\end{pmatrix}\) c) \(k:\vec{x}=\begin{pmatrix}3\\2\\1\end{pmatrix}+t\begin{pmatrix}-1\\2\\-1\end{pmatrix}\)

- Three lines form a triangle when each pair intersects at a different point. - Find the three pairwise intersections. - Verify that the intersection points are distinct. - Use the parameter values at the endpoints to determine each side's interval.

1. Equating \(g\) and \(h\) gives \(\lambda=2\) and \(\mu=0\), so their intersection is \(P_1=(5, 0, 1)\). 2. Equating \(h\) and \(k\) gives \(\mu=2\) and \(\nu=0\), so their intersection is \(P_2=(3, 4, 5)\). 3. Equating \(k\) and \(g\) gives \(\nu=-2\) and \(\lambda=0\), so their intersection is \(P_3=(1, 2, 3)\). 4. The three pairwise intersection points are distinct, so the lines form a triangle. 5. The side on \(g\) runs from \(\lambda=0\) to \(\lambda=2\). The side on \(h\) runs from \(\mu=0\) to \(\mu=2\). The side on \(k\) runs from \(\nu=-2\) to \(\nu=0\).

The lines form a triangle with vertices \((5, 0, 1)\), \((3, 4, 5)\), and \((1, 2, 3)\). The parameter intervals are \(\lambda\in[0, 2]\) on \(g\), \(\mu\in[0, 2]\) on \(h\), and \(\nu\in[-2, 0]\) on \(k\).

- A triangle requires three distinct pairwise intersections. - Check the relationship of each pair of lines. - What does one skew pair imply about the possibility of a triangle?

1. Equating \(g\) and \(h\) gives \(\lambda=1\) and \(\mu=0\), and the third coordinate is consistent. Thus they intersect at \((3, 1, 2)\). 2. Equating \(g\) and \(k\) gives \(\nu=0\) from the second coordinate and \(\lambda=0\) from the third coordinate. The first coordinate would then require \(2=1\), a contradiction. 3. The direction vectors of \(g\) and \(k\) are not scalar multiples, so these lines are skew. 4. Because one pair of lines does not intersect, the three lines do not form a triangle.

The lines do not form a triangle because \(g\) and \(k\) are skew.

- A point reflected through \(Q\) has vector \(2\vec{q}-\vec{x}\). - Reflect one position point from the line. - What happens to a displacement vector under point reflection? - Use the general point \(\vec{a}+s\vec{v}\) for part c.

1. Reflect the position point \(\vec{a}=\begin{pmatrix}-1\\2\\5\end{pmatrix}\) through \(Q\): \(\vec{a'}=2\vec{q}-\vec{a}=\begin{pmatrix}3\\0\\-1\end{pmatrix}\). 2. Point reflection reverses a direction vector, so the image direction is \(-\vec{v}\). An equivalent equation may use either \(\vec{v}\) or \(-\vec{v}\). One equation is \(g':\vec{x}=\begin{pmatrix}3\\0\\-1\end{pmatrix}+s\begin{pmatrix}2\\0\\-1\end{pmatrix}\). 3. The two lines have parallel directions. Because \(Q\notin g\), the reflection cannot map \(g\) onto itself, so they are distinct and parallel. 4. In general, a point \(\vec{x}=\vec{a}+s\vec{v}\) maps to \(\vec{x'}=2\vec{q}-\vec{x}=(2\vec{q}-\vec{a})-s\vec{v}\). The image direction \(-\vec{v}\) is parallel to \(\vec{v}\).

a) One possible equation is \(g':\vec{x}=\begin{pmatrix}3\\0\\-1\end{pmatrix}+s\begin{pmatrix}2\\0\\-1\end{pmatrix}\). b) The lines are distinct and parallel. c) Reflection changes the direction vector from \(\vec{v}\) to \(-\vec{v}\), which is parallel to \(\vec{v}\).

- Equate corresponding coordinates. - First solve the equations that do not contain \(a\). - Use those parameter values in the remaining equation to find \(a\). - Substitute a line parameter to calculate the common point.

1. Equating coordinates gives \(4+2r=a+s\), \(2-r=1+2s\), and \(1+3r=5-s\). 2. Solving the second and third equations gives \(r=\frac{7}{5}\) and \(s=-\frac{1}{5}\). 3. Substitution into the first equation gives \(4+2\cdot\frac{7}{5}=a-\frac{1}{5}\), so \(a=7\). 4. Substituting \(r=\frac{7}{5}\) into \(g\) gives the intersection point \(\left(\frac{34}{5},\frac{3}{5},\frac{26}{5}\right)\).

The lines intersect when \(a=7\), at \(\left(\frac{34}{5}, \frac{3}{5}, \frac{26}{5}\right)\).

- Find a direction vector for the line through \(A\) and \(B\). - Compare the two direction vectors. - If they are parallel, test whether one line's point lies on the other. - Inconsistent parameter values in a point test rule out identical lines.

1. A direction vector for \(g\) is \(\overrightarrow{AB}=\begin{pmatrix}2\\2\\-1\end{pmatrix}\), so \(g:\vec{x}=\begin{pmatrix}2\\-1\\4\end{pmatrix}+r\begin{pmatrix}2\\2\\-1\end{pmatrix}\). 2. The direction vector of \(g\) is \(2\) times the direction vector of \(h\), so the lines are parallel or identical. 3. Testing point \(A\) in \(h\), the first coordinate gives \(t=1\), while the second gives \(t=-1\). The point is not on \(h\). 4. Therefore, the lines are distinct and parallel.

The lines are distinct and parallel.

- Compare the direction vectors first. - For parallel directions, use a point test. - For nonparallel directions, equate coordinates to find a possible intersection. - If two lines are identical, any intersection with one is also an intersection with the other.

1. The direction vector of \(k\) is \(-2\) times the direction vector of \(g\). The position point \((3, 2, -1)\) of \(k\) lies on \(g\) when \(r=2\). Therefore, \(g\) and \(k\) are identical. 2. The direction vectors of \(g\) and \(h\) are not scalar multiples. Equating coordinates gives \(r=1\) and \(s=1\), and all three coordinate equations are satisfied. 3. Substituting \(r=1\) into \(g\) gives the intersection point \((1, 1, 1)\). Since \(g=k\), line \(h\) also intersects \(k\) at this point.

Lines \(g\) and \(k\) are identical. Line \(h\) intersects both at \((1, 1, 1)\).

- Begin with coordinate equations that do not contain \(a\). - Use those parameter values in the remaining equation. - What relationship applies to nonparallel lines with no common point?

1. The direction vectors are not scalar multiples, so the lines are never parallel or identical. 2. Equating coordinates gives \(2+\lambda=2\mu\), \(-1+2\lambda=a+\mu\), and \(3=1+2\mu\). 3. The third equation gives \(\mu=1\), and the first then gives \(\lambda=0\). 4. The second equation gives \(-1=a+1\), so \(a=-2\). The common point is the position point of \(g\), \((2, -1, 3)\). 5. For \(a\neq-2\), the lines are nonparallel and do not intersect, so they are skew.

For \(a=-2\), the lines intersect at \((2, -1, 3)\). For \(a\neq-2\), the lines are skew.

- How can you find a direction vector from two points? - What must be true about the parameter for a point to lie on the line? - Which coordinate is always \(0\) in the \(xy\)-plane? - What do parallel lines have in common about their direction vectors?

1. A direction vector is \(\overrightarrow{AB}=\begin{pmatrix}-3\\3\\-3\end{pmatrix}\). Therefore, \(g:\vec{x}=\begin{pmatrix}4\\-1\\2\end{pmatrix}+r\begin{pmatrix}-3\\3\\-3\end{pmatrix}\). 2. For \(C\), all three coordinate equations give \(r=2\), so \(C\) lies on \(g\). For \(D\), the \(x\)- and \(y\)-coordinates give \(r=-1\), but the resulting \(z\)-coordinate is \(5\ne6\). Thus \(D\) does not lie on \(g\). 3. In the \(xy\)-plane, \(z=0\). Solve \(2-3r=0\), so \(r=\frac{2}{3}\). Then \(x=4-3\cdot\frac{2}{3}=2\) and \(y=-1+3\cdot\frac{2}{3}=1\). The intersection point is \((2, 1, 0)\). 4. A parallel line can use the same direction vector. Thus \(h:\vec{x}=\begin{pmatrix}0\\0\\5\end{pmatrix}+s\begin{pmatrix}-3\\3\\-3\end{pmatrix}\).

a) \(g:\vec{x}=\begin{pmatrix}4\\-1\\2\end{pmatrix}+r\begin{pmatrix}-3\\3\\-3\end{pmatrix}\) b) Point \(C\) lies on \(g\); point \(D\) does not. c) \((2, 1, 0)\) d) \(h:\vec{x}=\begin{pmatrix}0\\0\\5\end{pmatrix}+s\begin{pmatrix}-3\\3\\-3\end{pmatrix}\)

- Compare the direction vectors for each pair. - For parallel directions, use a point test. - For nonparallel directions, solve for possible intersection parameters. - Verify every coordinate equation before accepting an intersection.

1. The direction vector of \(h\) is \(-1.5\) times the direction vector of \(g\). Testing the point \((0, 1, -5)\) from \(h\) on \(g\) gives \(r=-2\) from the first coordinate, but then the second coordinate is \(2\), not \(1\). Thus \(g\) and \(h\) are distinct and parallel. 2. The direction vectors of \(g\) and \(k\) are not scalar multiples. Equating coordinates gives \(r=-\frac{3}{2}\) and \(t=-2\), and all coordinates agree. Their intersection point is \((1, 1, -5)\). 3. The direction vectors of \(h\) and \(k\) are not scalar multiples. The first two coordinate equations give \(s=0\) and \(t=-3\), but the third coordinates are \(-5\) and \(-6\). Therefore, \(h\) and \(k\) are skew.

\(g\) and \(h\) are distinct and parallel. \(g\) and \(k\) intersect at \((1, 1, -5)\). \(h\) and \(k\) are skew.

- Use one known point as the position vector. - Form two connecting vectors from the given points. - Find a normal vector by requiring zero dot products with both spanning vectors. - Use the normal-vector components as the coefficients in Cartesian form. - Substitute a known point to determine the constant.

1. Use \(A\) as a point and form the spanning vectors \(\overrightarrow{AB}=\begin{pmatrix}2\\0\\-2\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}0\\3\\-2\end{pmatrix}\). Thus \(E:\vec{x}=\begin{pmatrix}1\\2\\3\end{pmatrix}+r\begin{pmatrix}2\\0\\-2\end{pmatrix}+s\begin{pmatrix}0\\3\\-2\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(2a-2c=0\) and \(3b-2c=0\). One solution is \(\vec{n}=\begin{pmatrix}3\\2\\3\end{pmatrix}\). 3. Using point \(A\), the Cartesian equation is \(E:3x+2y+3z=16\).

a) \(E:\vec{x}=\begin{pmatrix}1\\2\\3\end{pmatrix}+r\begin{pmatrix}2\\0\\-2\end{pmatrix}+s\begin{pmatrix}0\\3\\-2\end{pmatrix}\) b) \(E:3x+2y+3z=16\)

- Form two vectors connecting one vertex to the other two. - Find a normal vector by requiring zero dot products with both spanning vectors. - Convert point-normal form to Cartesian form for an easy point test. - A point in the plane must satisfy its equation.

1. Two spanning vectors are \(\overrightarrow{PQ}=\begin{pmatrix}4\\0\\-4\end{pmatrix}\) and \(\overrightarrow{PR}=\begin{pmatrix}0\\4\\-4\end{pmatrix}\). Thus \(E:\vec{x}=\begin{pmatrix}2\\2\\6\end{pmatrix}+r\begin{pmatrix}4\\0\\-4\end{pmatrix}+s\begin{pmatrix}0\\4\\-4\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(4a-4c=0\) and \(4b-4c=0\), so a simplified normal vector is \(\begin{pmatrix}1\\1\\1\end{pmatrix}\). A point-normal equation is \(\left(\vec{x}-\begin{pmatrix}2\\2\\6\end{pmatrix}\right)\cdot\begin{pmatrix}1\\1\\1\end{pmatrix}=0\). 3. The corresponding Cartesian equation is \(x+y+z=10\). Since \(3+3+4=10\), the sensor point lies in the plane.

a) \(E:\vec{x}=\begin{pmatrix}2\\2\\6\end{pmatrix}+r\begin{pmatrix}4\\0\\-4\end{pmatrix}+s\begin{pmatrix}0\\4\\-4\end{pmatrix}\) b) \(\left(\vec{x}-\begin{pmatrix}2\\2\\6\end{pmatrix}\right)\cdot\begin{pmatrix}1\\1\\1\end{pmatrix}=0\) c) Yes, \(S=(3, 3, 4)\) lies in \(E\).

- Form two connecting vectors from point \(A\). - When do two direction vectors fail to span a plane? - What does dependence of the two connecting vectors mean for the three points?

1. The connecting vectors are \(\overrightarrow{AB}=\begin{pmatrix}2\\1\\2\end{pmatrix}\) and \(\overrightarrow{AC_k}=\begin{pmatrix}4\\2\\k-1\end{pmatrix}\). Thus the point set is \(\vec{x}=\begin{pmatrix}1\\2\\1\end{pmatrix}+r\begin{pmatrix}2\\1\\2\end{pmatrix}+s\begin{pmatrix}4\\2\\k-1\end{pmatrix}\). 2. The points fail to determine a plane when the two direction vectors are scalar multiples. The first two coordinates require \(\overrightarrow{AC_k}=2\overrightarrow{AB}\). Therefore, \(k-1=2\cdot2=4\), so \(k=5\). 3. For \(k=5\), \(\overrightarrow{AC_5}=2\overrightarrow{AB}\), so the three points are collinear.

a) \(\vec{x}=\begin{pmatrix}1\\2\\1\end{pmatrix}+r\begin{pmatrix}2\\1\\2\end{pmatrix}+s\begin{pmatrix}4\\2\\k-1\end{pmatrix}\) b) \(k=5\) c) The three points are collinear.

- Use two known points to find a line direction. - Two intersecting lines with independent directions determine a plane. - Use their common point and both directions for the plane equation. - A point test must produce parameter values that satisfy all three coordinates.

1. The first line is \(h_1:\vec{x}=\begin{pmatrix}2\\1\\4\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\2\end{pmatrix}\). For the second line, \(\overrightarrow{PQ}=\begin{pmatrix}-1\\0\\-1\end{pmatrix}\), so \(h_2:\vec{x}=\begin{pmatrix}2\\1\\4\end{pmatrix}+\mu\begin{pmatrix}-1\\0\\-1\end{pmatrix}\). 2. The lines intersect at \(P\), and their direction vectors are not scalar multiples, so they determine a unique plane: \(E:\vec{x}=\begin{pmatrix}2\\1\\4\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\2\end{pmatrix}+\mu\begin{pmatrix}-1\\0\\-1\end{pmatrix}\). 3. The \(x\)-coordinate of \(R\) requires \(\mu=-1\), and the \(y\)-coordinate requires \(\lambda=1\). These values give \(z=4+2\cdot1-(-1)=7\), not \(6\). Therefore, \(R\notin E\).

a) \(h_1:\vec{x}=\begin{pmatrix}2\\1\\4\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\2\end{pmatrix}\); \(h_2:\vec{x}=\begin{pmatrix}2\\1\\4\end{pmatrix}+\mu\begin{pmatrix}-1\\0\\-1\end{pmatrix}\) b) \(E:\vec{x}=\begin{pmatrix}2\\1\\4\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\2\end{pmatrix}+\mu\begin{pmatrix}-1\\0\\-1\end{pmatrix}\) c) No, \(R=(3, 2, 6)\) does not lie in \(E\).

- A point on a line must satisfy all three coordinate equations for one parameter value. - Use the line direction and a vector from the line to the outside point as spanning vectors. - Find a normal vector by requiring zero dot products with both spanning vectors. - Use a known point to find the constant term.

1. The first coordinate would require \(s=2\), but every point on \(h\) has \(y=2\), while \(P\) has \(y=1\). Thus \(P\notin h\). 2. Use the line direction \(\vec{u}=\begin{pmatrix}1\\0\\2\end{pmatrix}\) and the vector from the line's given point to \(P\), \(\vec{v}=\begin{pmatrix}2\\-1\\4\end{pmatrix}\). Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(a+2c=0\) and \(2a-b+4c=0\). One solution is \(\vec{n}=\begin{pmatrix}2\\0\\-1\end{pmatrix}\). 3. Using \((3, 2, -1)\), the Cartesian equation is \(E:2x-z=7\).

a) \(P\notin h\). b) \(E:2x-z=7\)

- Three points determine a unique plane when they are not collinear. - Use one point and two connecting vectors for parametric form. - A point on the \(z\)-axis has \(x=0\) and \(y=0\). - Solve the resulting two equations for the parameters.

1. \(\overrightarrow{AB}=\begin{pmatrix}-3\\3\\1\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}-2\\-1\\2\end{pmatrix}\) are not scalar multiples, so the points are not collinear and determine a unique plane. 2. One equation is \(E:\vec{x}=\begin{pmatrix}4\\2\\1\end{pmatrix}+r\begin{pmatrix}-3\\3\\1\end{pmatrix}+s\begin{pmatrix}-2\\-1\\2\end{pmatrix}\). 3. On the \(z\)-axis, \(x=0\) and \(y=0\). Solving \(4-3r-2s=0\) and \(2+3r-s=0\) gives \(r=0\), \(s=2\). Then \(z=1+r+2s=5\), so the intersection point is \((0, 0, 5)\).

a) The two connecting vectors are not scalar multiples. b) \(E:\vec{x}=\begin{pmatrix}4\\2\\1\end{pmatrix}+r\begin{pmatrix}-3\\3\\1\end{pmatrix}+s\begin{pmatrix}-2\\-1\\2\end{pmatrix}\) c) \((0, 0, 5)\)

- Test whether one parameter value can match all three point coordinates. - Use the line direction as one spanning vector. - Use a vector from the line to the additional point as the second spanning vector. - Find the plane normal by requiring zero dot products with both spanning vectors. - The normal-vector components become the coefficients in the Cartesian equation.

1. Every point on \(g\) has \(y=2\), while \(P\) has \(y=4\). Therefore, \(P\notin g\). 2. Use the line direction \(\begin{pmatrix}2\\0\\-1\end{pmatrix}\) and the vector from the line's given point to \(P\), \(\begin{pmatrix}4\\2\\-2\end{pmatrix}\). Thus \(E:\vec{x}=\begin{pmatrix}1\\2\\3\end{pmatrix}+r\begin{pmatrix}2\\0\\-1\end{pmatrix}+s\begin{pmatrix}4\\2\\-2\end{pmatrix}\). 3. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(2a-c=0\) and \(4a+2b-2c=0\). One solution is \(\vec{n}=\begin{pmatrix}1\\0\\2\end{pmatrix}\). Using \(P\), the Cartesian equation is \(E:x+2z=7\).

a) \(P\notin g\). b) \(E:\vec{x}=\begin{pmatrix}1\\2\\3\end{pmatrix}+r\begin{pmatrix}2\\0\\-1\end{pmatrix}+s\begin{pmatrix}4\\2\\-2\end{pmatrix}\) c) \(E:x+2z=7\)

- Parallel direction vectors are scalar multiples. - Test whether a point from one line lies on the other to distinguish identical lines from distinct parallel lines. - Use a line direction and a vector connecting the two lines as spanning vectors. - A point test must satisfy all three coordinates.

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so the lines are parallel. The point \((1, 3, 2)\) is not on \(g\): the \(x\)-coordinate would require \(r=-1\), which gives \(y=-1\), not \(3\). Thus the lines are distinct. 2. Use the direction \(\begin{pmatrix}1\\1\\-1\end{pmatrix}\) and the vector between the given points, \(\begin{pmatrix}-1\\3\\1\end{pmatrix}\). One equation is \(E:\vec{x}=\begin{pmatrix}2\\0\\1\end{pmatrix}+r\begin{pmatrix}1\\1\\-1\end{pmatrix}+t\begin{pmatrix}-1\\3\\1\end{pmatrix}\). 3. Setting this expression equal to \(P\) gives \(r=\frac{1}{4}\) and \(t=\frac{5}{4}\), which satisfy all three coordinates. Therefore, \(P\in E\).

a) The lines are distinct and parallel. b) \(E:\vec{x}=\begin{pmatrix}2\\0\\1\end{pmatrix}+r\begin{pmatrix}1\\1\\-1\end{pmatrix}+t\begin{pmatrix}-1\\3\\1\end{pmatrix}\) c) Yes. The values \(r=\frac{1}{4}\) and \(t=\frac{5}{4}\) produce \(P\).

- Distinct parallel lines determine exactly one plane. - Parallel direction vectors are linearly dependent. - Use the common direction and a vector joining the lines as spanning vectors. - Find a normal vector by requiring zero dot products with both spanning vectors.

1. The direction vectors must be nonzero scalar multiples, and \(\vec{q}-\vec{p}\) must not be a scalar multiple of the common direction. These conditions make the lines distinct and parallel. 2. Use \(\vec{u}=\begin{pmatrix}2\\1\\0\end{pmatrix}\) and \(\vec{q}-\vec{p}=\begin{pmatrix}-1\\3\\1\end{pmatrix}\) as spanning vectors. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(2a+b=0\) and \(-a+3b+c=0\). One solution is \(\vec{n}=\begin{pmatrix}1\\-2\\7\end{pmatrix}\). 3. Using \((1, -1, 2)\), the plane equation is \(E:x-2y+7z=17\).

a) The direction vectors must be nonzero scalar multiples, and the vector between the given points must not be parallel to them. b) \(E:x-2y+7z=17\)

- Two lines define a plane when they intersect or are distinct and parallel. - Compare direction vectors first. - If the lines are not parallel, solve for a possible intersection. - Skew lines do not lie in a common plane.

1. In part a, the direction vectors are not scalar multiples. Equating coordinates gives \(\lambda=1\) and \(\mu=0\), and all three coordinates agree at \(S=(4, 2, 2)\). The intersecting lines determine a unique plane. 2. In part b, the direction vectors are not scalar multiples. Equating the \(x\)-coordinates gives \(\lambda=0\), and then the \(y\)-coordinates give \(\mu=0\). The \(z\)-coordinates would require \(3=4\), a contradiction. Thus the lines are skew and do not lie in one plane.

a) Yes. The lines intersect at \(S=(4, 2, 2)\). b) No. The lines are skew.

- Equate the point's coordinates with the three coordinate expressions of the plane. - Solve two equations for \(r\) and \(s\). - What does it mean if the resulting values fail to satisfy the third equation? - Always check the unused coordinate equation.

1. For \(P\), equating coordinates gives \(2r+s=3\), \(-r+3s=2\), and \(4r-2s=2\). The first two equations give \(r=1\) and \(s=1\), and these values satisfy the third equation. Therefore, \(P\) lies in \(E\). 2. For \(Q\), the coordinate equations are \(2r+s=-1\), \(-r+3s=1\), and \(4r-2s=3\). The first two equations give \(r=-\frac{4}{7}\) and \(s=\frac{1}{7}\). Then \(4r-2s=-\frac{18}{7}\ne3\), so no single pair of parameter values satisfies all three equations. Therefore, \(Q\) does not lie in \(E\).

a) Point \(P\) lies in \(E\) for \(r=1\) and \(s=1\). b) Point \(Q\) does not lie in \(E\).

- Write a parametric equation through each pair of points. - Compare the direction vectors. - If the lines are not parallel, test whether they intersect. - What does a pair of skew lines imply about coplanarity?

1. Parametric equations are \(g:\vec{x}=\begin{pmatrix}1\\2\\1\end{pmatrix}+r\begin{pmatrix}2\\2\\2\end{pmatrix}\) and \(h:\vec{x}=\begin{pmatrix}2\\0\\4\end{pmatrix}+s\begin{pmatrix}2\\1\\1\end{pmatrix}\). 2. The direction vectors are not scalar multiples, so the lines are not parallel. 3. Equating coordinates gives a system with no solution. For example, the first two equations give \(s=-3\) and \(r=-\frac{5}{2}\), but these values fail the third equation. 4. Thus the lines are skew. Skew lines are not coplanar, so the four points do not lie in one plane.

The lines are skew. Therefore, the four points do not lie in a common plane.

- Three noncollinear points determine a plane. - Write a parametric equation using three of the points. - Test whether the fourth point can be produced by one pair of parameter values. - The same parameters must satisfy all three coordinates.

1. The plane through \(A\), \(B\), and \(C\) can be written as \(E:\vec{x}=\begin{pmatrix}2\\1\\3\end{pmatrix}+r\begin{pmatrix}3\\1\\-2\end{pmatrix}+s\begin{pmatrix}-1\\3\\-1\end{pmatrix}\). 2. Set this expression equal to \(D\). The resulting system is \(3r-s=5\), \(r+3s=5\), and \(2r+s=5\). 3. The system has the solution \(r=2\), \(s=1\), which satisfies all three equations. Therefore, \(D\) lies in the plane through \(A\), \(B\), and \(C\), so all four points are coplanar.

Yes. All four points are coplanar; \(D\) corresponds to \(r=2\) and \(s=1\) in the plane through \(A\), \(B\), and \(C\).

- Set corresponding line coordinates equal to find an intersection. - Use the intersection point and both line directions for parametric form. - Find a plane normal by requiring zero dot products with both direction vectors. - Use the normal and a point to write Cartesian form.

1. Equating coordinates gives the system \(3+t=2+2s\), \(4+2t=2+s\), and \(1-t=2+s\). Solving gives \(t=-1\) and \(s=0\), so \(S=(2, 2, 2)\). 2. The intersection point and the two line directions give \(E:\vec{x}=\begin{pmatrix}2\\2\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\-1\end{pmatrix}+s\begin{pmatrix}2\\1\\1\end{pmatrix}\). 3. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both direction vectors. Then \(a+2b-c=0\) and \(2a+b+c=0\). One solution is \(\vec{n}=\begin{pmatrix}1\\-1\\-1\end{pmatrix}\). Using \(S\), the Cartesian equation is \(E:x-y-z=-2\).

1. \(S=(2, 2, 2)\) 2. \(E:\vec{x}=\begin{pmatrix}2\\2\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\-1\end{pmatrix}+s\begin{pmatrix}2\\1\\1\end{pmatrix}\) 3. \(E:x-y-z=-2\)

- Parallel direction vectors are scalar multiples. - Test a point from one line on the other to confirm the lines are distinct. - Use one line direction and a connecting vector as spanning vectors. - Find a normal vector by requiring zero dot products with both spanning vectors.

1. The first direction vector is \(-2\) times the second, so the lines are parallel. The point \((0, 1, 0)\) is not on \(g_1\): the \(x\)-coordinate would require \(r=-\frac{1}{2}\), which gives \(y=3\), not \(1\). Thus the lines are distinct. 2. Use direction \(\vec{u}=\begin{pmatrix}-1\\1\\-2\end{pmatrix}\) and the vector between the given points, \(\vec{v}=\begin{pmatrix}-1\\-1\\-3\end{pmatrix}\). One equation is \(E:\vec{x}=\begin{pmatrix}0\\1\\0\end{pmatrix}+s\begin{pmatrix}-1\\1\\-2\end{pmatrix}+k\begin{pmatrix}-1\\-1\\-3\end{pmatrix}\). 3. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(-a+b-2c=0\) and \(-a-b-3c=0\). One solution is \(\vec{n}=\begin{pmatrix}-5\\-1\\2\end{pmatrix}\). Using \((0, 1, 0)\) gives \(-5x-y+2z=-1\), or equivalently \(E:5x+y-2z=1\).

1. The lines are distinct and parallel. 2. \(E:\vec{x}=\begin{pmatrix}0\\1\\0\end{pmatrix}+s\begin{pmatrix}-1\\1\\-2\end{pmatrix}+k\begin{pmatrix}-1\\-1\\-3\end{pmatrix}\) 3. \(E:5x+y-2z=1\)

- Use the line direction and a vector from the line to the additional point. - Confirm that the spanning vectors are not parallel. - Find a normal vector by requiring zero dot products with both spanning vectors. - Use the normal-vector components as coefficients in the Cartesian equation. - Test the point in the Cartesian equation.

1. Use the line direction \(\vec{u}=\begin{pmatrix}1\\0\\1\end{pmatrix}\) and the vector from the line's given point to \(Q\), \(\vec{v}=\begin{pmatrix}1\\-3\\3\end{pmatrix}\). Thus \(E:\vec{x}=\begin{pmatrix}1\\2\\0\end{pmatrix}+r\begin{pmatrix}1\\0\\1\end{pmatrix}+s\begin{pmatrix}1\\-3\\3\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(a+c=0\) and \(a-3b+3c=0\). One solution is \(\vec{n}=\begin{pmatrix}3\\-2\\-3\end{pmatrix}\). Using \((1, 2, 0)\), the Cartesian equation is \(E:3x-2y-3z=-1\). 3. For \(P\), \(3\cdot4-2\cdot(-1)-3\cdot5=-1\). Therefore, \(P\in E\).

1. \(E:\vec{x}=\begin{pmatrix}1\\2\\0\end{pmatrix}+r\begin{pmatrix}1\\0\\1\end{pmatrix}+s\begin{pmatrix}1\\-3\\3\end{pmatrix}\) 2. \(E:3x-2y-3z=-1\) 3. Yes, \(P=(4, -1, 5)\) lies in \(E\).

- Parallel direction vectors are scalar multiples. - Test whether a point from one line lies on the other. - Use a direction vector and a vector joining the lines as spanning vectors. - Find a plane normal by requiring zero dot products with both spanning vectors.

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so the lines are parallel. Testing \((0, 1, 2)\) on \(h\) gives inconsistent parameter values from the three coordinates, so the lines are distinct. 2. Use the line direction \(\vec{v}=\begin{pmatrix}2\\-1\\1\end{pmatrix}\) and the vector between the given points, \(\vec{u}=\begin{pmatrix}3\\-1\\2\end{pmatrix}\). Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(2a-b+c=0\) and \(3a-b+2c=0\). One solution is \(\vec{n}=\begin{pmatrix}-1\\-1\\1\end{pmatrix}\). 3. Using \((0, 1, 2)\), the plane equation is \(E:-x-y+z=1\).

1. The lines are distinct and parallel. 2. \(E:-x-y+z=1\)

- Use one point and two connecting vectors for parametric form. - Find a normal vector by requiring zero dot products with both spanning vectors. - Use a known point to determine the constant in Cartesian form. - Substitute the coordinates of \(D\) and solve for \(z\).

1. Two spanning vectors are \(\overrightarrow{AB}=\begin{pmatrix}-2\\2\\-1\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}-4\\0\\4\end{pmatrix}\). Thus \(E:\vec{x}=\begin{pmatrix}4\\1\\1\end{pmatrix}+r\begin{pmatrix}-2\\2\\-1\end{pmatrix}+s\begin{pmatrix}-4\\0\\4\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(-2a+2b-c=0\) and \(-4a+4c=0\). One solution is \(\vec{n}=\begin{pmatrix}2\\3\\2\end{pmatrix}\). Using \(A\), the Cartesian equation is \(E:2x+3y+2z=13\). 3. Substitute \(D\): \(2\cdot1+3\cdot2+2z=13\). Thus \(8+2z=13\), so \(z=\frac{5}{2}=2.5\).

a) \(E:\vec{x}=\begin{pmatrix}4\\1\\1\end{pmatrix}+r\begin{pmatrix}-2\\2\\-1\end{pmatrix}+s\begin{pmatrix}-4\\0\\4\end{pmatrix}\) b) \(E:2x+3y+2z=13\) c) \(z=2.5\)

- A line lies in a plane when substitution gives a true equation for every parameter value. - A plane normal must be perpendicular to every direction in the plane. - Make the second plane distinct by choosing a normal not parallel to the first plane's normal. - Use a point on the line to find the constant term.

1. Substitute the line coordinates into \(E_1\): \(2(2+s)-2s=4\). This identity holds for every \(s\), so \(g\subset E_1\). 2. A normal vector to \(E_2\) must be perpendicular to the line direction \(\begin{pmatrix}1\\2\\-1\end{pmatrix}\). Choose \(\vec{n}_2=\begin{pmatrix}1\\0\\1\end{pmatrix}\), since their dot product is \(0\). 3. This normal is not a scalar multiple of the normal \(\begin{pmatrix}2\\-1\\0\end{pmatrix}\) of \(E_1\), so the planes are distinct and intersect in a line. 4. Using the point \((2, 0, 1)\), one possible equation is \(E_2:x+z=3\). Both planes contain \(g\), so their intersection line is \(g\).

The substitution gives an identity, so \(g\subset E_1\). One possible second plane is \(E_2:x+z=3\).

- What happens when one parameter is fixed and the other varies over an interval? - What region is generated when two independent parameters each vary from \(0\) to \(1\)? - Which parameter pairs are the boundary vertices of the triangular restriction? - Substitute \(s=r\) and combine the direction vectors.

1. Fixing \(r=2\) gives one fixed starting point, while \(s\) varies over a closed interval. This describes a line segment. 2. When \(r\) and \(s\) vary independently from \(0\) to \(1\), the two direction vectors generate a parallelogram. 3. The restrictions \(r\ge0\), \(s\ge0\), and \(r+s\le1\) describe the triangular region with parameter vertices \((0, 0)\), \((1, 0)\), and \((0, 1)\). 4. Substituting \(s=r\) gives \(\vec{x}=\begin{pmatrix}3\\0\\-2\end{pmatrix}+r\begin{pmatrix}1\\3\\2\end{pmatrix}\), which describes a line. 5. For part 2, the parameter pairs \((0, 0)\), \((1, 0)\), \((0, 1)\), and \((1, 1)\) give the vertices \((3, 0, -2)\), \((5, 1, 0)\), \((2, 2, -2)\), and \((4, 3, 0)\).

1. A line segment 2. A parallelogram with vertices \((3, 0, -2)\), \((5, 1, 0)\), \((2, 2, -2)\), and \((4, 3, 0)\) 3. A triangle 4. A line

- Identify which parameters are fixed, bounded, or unrestricted. - An unrestricted parameter extends a figure infinitely in its direction. - Use the parameter values associated with each listed vertex. - A restriction on \(\lambda+\mu\) commonly forms the slanted boundary of a triangle.

1. Figure (A) requires independent parameter intervals reaching \(2\) in the \(\vec{u}\)-direction and \(3\) in the \(\vec{v}\)-direction, so it matches restriction 2. 2. For figure (B), fixing \(\lambda=1\) and allowing \(\mu\) to vary gives a line parallel to \(\vec{v}\). Because \(\lambda\ne0\), it does not pass through \(\vec{a}\). Thus it matches restriction 3. 3. Figure (C) is the standard triangular parameter region \(\lambda\ge0\), \(\mu\ge0\), and \(\lambda+\mu\le1\), so it matches restriction 1. 4. In restriction 4, \(\lambda\) stays between two values while \(\mu\) is unrestricted. This produces an infinite strip parallel to \(\vec{v}\), so figure (D) matches restriction 4.

(A)–2, (B)–3, (C)–1, (D)–4

- Compare two connecting vectors for parallelism. - Use one point and two independent connecting vectors for parametric form. - Find a normal vector by requiring zero dot products with both spanning vectors. - A point on the \(x\)-axis has \(y=0\) and \(z=0\).

1. The vectors \(\overrightarrow{AB}=\begin{pmatrix}3\\-1\\-2\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}-1\\3\\-1\end{pmatrix}\) are not scalar multiples, so the points are not collinear. 2. One parametric equation is \(E:\vec{x}=\begin{pmatrix}2\\1\\3\end{pmatrix}+r\begin{pmatrix}3\\-1\\-2\end{pmatrix}+s\begin{pmatrix}-1\\3\\-1\end{pmatrix}\). 3. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(3a-b-2c=0\) and \(-a+3b-c=0\). One solution is \(\vec{n}=\begin{pmatrix}7\\5\\8\end{pmatrix}\). Using \(A\), the Cartesian equation is \(E:7x+5y+8z=43\). 4. On the \(x\)-axis, \(y=z=0\), so \(7x=43\). The intersection point is \(\left(\frac{43}{7}, 0, 0\right)\).

a) The two connecting vectors are not scalar multiples. b) \(E:\vec{x}=\begin{pmatrix}2\\1\\3\end{pmatrix}+r\begin{pmatrix}3\\-1\\-2\end{pmatrix}+s\begin{pmatrix}-1\\3\\-1\end{pmatrix}\) c) \(E:7x+5y+8z=43\) d) \(\left(\frac{43}{7}, 0, 0\right)\)

- A point-on-line test must give one parameter value that works for every coordinate. - Use the line direction and a vector from the line to the outside point as spanning vectors. - Find a normal vector by requiring zero dot products with both spanning vectors. - Test \(R\) in the Cartesian equation.

1. Matching the \(x\)-coordinate of \(P\) would require \(t=-3\), while matching the \(y\)-coordinate would require \(t=2\). Therefore, \(P\notin h\). 2. Use the line direction \(\vec{u}=\begin{pmatrix}1\\1\\3\end{pmatrix}\) and the vector from the line's given point to \(P\), \(\vec{v}=\begin{pmatrix}-3\\2\\3\end{pmatrix}\). Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(a+b+3c=0\) and \(-3a+2b+3c=0\). One solution is \(\vec{n}=\begin{pmatrix}-3\\-12\\5\end{pmatrix}\). 3. Using \((4, 0, -2)\), the plane equation is \(-3x-12y+5z=-22\), or equivalently \(F:3x+12y-5z=22\). 4. For \(R\), \(3\cdot2+12\cdot3-5\cdot4=22\). Therefore, \(R\in F\).

a) \(P\notin h\). b) \(F:3x+12y-5z=22\) c) Yes, \(R=(2, 3, 4)\) lies in \(F\).

- How far is each side of the square from its center? - Write a parametric equation using endpoints \(S\) and \(A\). - For a point on the segment, what range must the parameter satisfy?

1. The base is centered at the origin and has side length \(8\), so each base coordinate is \(4\) units from the corresponding axis. One consistent labeling is \(A(4, -4, 0)\), \(B(4, 4, 0)\), \(C(-4, 4, 0)\), and \(D(-4, -4, 0)\). 2. A parametric equation of the line through \(S\) and \(A\) is \(\vec{x}=\begin{pmatrix}0\\0\\12\end{pmatrix}+t\begin{pmatrix}4\\-4\\-12\end{pmatrix}\). The edge corresponds to \(0\le t\le1\). 3. Substituting \(P\) gives \(2=4t\), \(-2=-4t\), and \(6=12-12t\). Each equation gives \(t=\frac{1}{2}\). 4. Because \(\frac{1}{2}\) is between \(0\) and \(1\), point \(P\) lies on edge \(\overline{SA}\).

a) One consistent labeling is \(A(4, -4, 0)\), \(B(4, 4, 0)\), \(C(-4, 4, 0)\), and \(D(-4, -4, 0)\). b) Yes. Point \(P(2, -2, 6)\) lies on \(\overline{SA}\) when \(t=\frac{1}{2}\).

- First determine whether the direction vectors are scalar multiples. - How can you test whether two nonparallel lines have a common point? - How does multiplying a vector by a scalar change its magnitude? - What coordinates must a point on the \(x\)-axis have? - Two lines intersect when their coordinate equations have a common solution.

1. The direction vectors \(\begin{pmatrix}2\\4\\-4\end{pmatrix}\) and \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) are not scalar multiples, so the lines are not parallel. Equating the coordinates gives \(4+2t=1+s\), \(-1+4t=3+s\), and \(2-4t=s\). The first two equations give \(t=\frac{7}{2}\) and \(s=10\), but the third would require \(-12=10\). The lines do not intersect, so they are skew. 2. The direction vector of \(g\) has magnitude \(\sqrt{2^2+4^2+(-4)^2}=6\). Scaling by \(\frac{1}{2}\) gives \(\begin{pmatrix}1\\2\\-2\end{pmatrix}\), which has magnitude \(3\). 3. Write \(n\) as \(\vec{x}=\begin{pmatrix}p\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\1\end{pmatrix}\). Equating it with \(g\) gives \(p+\lambda=4+2t\), \(\lambda=-1+4t\), and \(\lambda=2-4t\). The last two equations give \(t=\frac{3}{8}\) and \(\lambda=\frac{1}{2}\). Then \(p+\frac{1}{2}=4+2\cdot\frac{3}{8}\), so \(p=\frac{17}{4}\). Thus \(P=\left(\frac{17}{4}, 0, 0\right)\).

a) Lines \(g\) and \(h\) are skew. b) One possible direction vector is \(\begin{pmatrix}1\\2\\-2\end{pmatrix}\). c) \(P=\left(\frac{17}{4}, 0, 0\right)\)

- First write the equation of the line through \(A\) and \(B\). - When are two vectors parallel, and how can you test this algebraically? - How do you scale a vector to a specified magnitude? - What coordinates characterize a point on the \(z\)-axis? - Express a general point on \(h\) using its parameter.

1. A direction vector for \(g\) is \(\overrightarrow{AB}=\begin{pmatrix}2\\0\\2\end{pmatrix}\), so use the equivalent vector \(\begin{pmatrix}1\\0\\1\end{pmatrix}\). It is not a scalar multiple of \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\), so the lines are not parallel. Equating the lines gives \(1+t=\mu\), \(2=1-\mu\), and \(-1+t=4\). The second equation gives \(\mu=-1\), and the first then gives \(t=-2\), but the third would require \(-3=4\). The lines do not intersect, so they are skew. 2. A direction vector parallel to \(h\) has the form \(c\begin{pmatrix}1\\-1\\0\end{pmatrix}\). Its magnitude is \(|c|\sqrt{2}\). Since \(\sqrt{18}=3\sqrt{2}\), choose \(c=3\). One possible vector is \(\begin{pmatrix}3\\-3\\0\end{pmatrix}\). 3. A point on \(h\) has coordinates \(C=(\mu, 1-\mu, 4)\). Then \(j\) can be written as \(\vec{x}=\begin{pmatrix}\mu\\1-\mu\\4\end{pmatrix}+\tau\begin{pmatrix}1\\0\\1\end{pmatrix}\). To intersect the \(z\)-axis, the \(x\)- and \(y\)-coordinates must both be \(0\). From \(1-\mu=0\), \(\mu=1\), and then \(1+\tau=0\), so \(\tau=-1\). Therefore, \(C=(1, 0, 4)\).

a) The direction vectors are not scalar multiples, and the coordinate equations have no common solution. Therefore, \(g\) and \(h\) are skew. b) One possible direction vector is \(\begin{pmatrix}3\\-3\\0\end{pmatrix}\). c) \(C=(1, 0, 4)\)

- Set the two expressions for \(\vec{x}\) equal. - How can you simplify when \(\vec{v}\) is a factor in every remaining term? - What does it mean geometrically when direction vectors are nonzero scalar multiples? - Which points are reached as a parameter ranges over all real numbers?

1. Set the two expressions equal: \(\vec{a}+t\vec{v}=(\vec{a}+2\vec{v})+s\left(-\frac{1}{2}\vec{v}\right)\). 2. Subtract \(\vec{a}\): \(t\vec{v}=2\vec{v}-\frac{1}{2}s\vec{v}\). Since \(\vec{v}\ne\vec{0}\), compare coefficients: \(t=2-\frac{1}{2}s\). Solving gives \(s=4-2t\). Thus every real \(t\) corresponds to a real \(s\). 3. More generally, let a new position vector be \(\vec{a}+r\vec{v}\), which is a point on the original line, and let the new direction vector be \(k\vec{v}\), where \(k\ne0\). The new equation gives \(\vec{a}+r\vec{v}+s(k\vec{v})=\vec{a}+(r+sk)\vec{v}\). 4. As \(s\) ranges over all real numbers, \(r+sk\) also ranges over all real numbers because \(k\ne0\). Therefore, the new equation produces exactly the same set of points.

a) \(s=4-2t\) b) Replacing the position vector by \(\vec{a}+r\vec{v}\) and the direction vector by \(k\vec{v}\), where \(k\ne0\), changes the parameter by \(t=r+sk\). This one-to-one correspondence covers all real values, so the point set is unchanged.

- Substitute the coordinates of \(P\) for \(\vec{x}\). - Treat each coordinate as a separate equation. - Which variable can be expressed directly in terms of \(\lambda\)? - Substitute that expression into another coordinate equation. - Could the resulting equation have more than one value of \(\lambda\)?

1. Equating coordinates gives \(7=a+2\lambda\), \(4=1+a\lambda\), and \(10=b+3\lambda\). 2. From the first equation, \(a=7-2\lambda\). Substitute this into the second equation: \(4=1+(7-2\lambda)\lambda\). Thus \(2\lambda^2-7\lambda+3=0\). 3. Factor: \((2\lambda-1)(\lambda-3)=0\). Therefore, \(\lambda=\frac{1}{2}\) or \(\lambda=3\). 4. If \(\lambda=3\), then \(a=7-2\cdot3=1\) and \(b=10-3\cdot3=1\). 5. If \(\lambda=\frac{1}{2}\), then \(a=7-2\cdot\frac{1}{2}=6\) and \(b=10-3\cdot\frac{1}{2}=\frac{17}{2}\).

\((a, b)=(1, 1)\) or \((a, b)=\left(6, \frac{17}{2}\right)\)