- What part of a parametrization determines the line’s direction? - A parallel line may use the same direction vector as the given line. - Use the required point as the location reached when the new parameter is zero.

1. A line parallel to \(g\) can use the same direction vector \((4, -3)\). 2. Using \(P=(5, 0)\) as the base point gives \(x=5+4t,\ y=-3t\).

One answer is \(x=5+4t,\ y=-3t\), where \(t\in\mathbb{R}\).

- The same parameter value is used in both coordinate equations. - Find the \(x\)-coordinate and \(y\)-coordinate separately before writing the ordered pair.

1. Substitute \(t=2\) into both coordinate equations: \(x=2(2)-1=3\) and \(y=2+3=5\). 2. The point is \((3, 5)\).

\((3, 5)\)

- Each parameter value gives one x-coordinate and one y-coordinate. - Keep the coordinates together as ordered pairs after each substitution.

1. For \(t=-1\), \(x=-2\) and \(y=2\). 2. For \(t=0\), \(x=0\) and \(y=1\). 3. For \(t=2\), \(x=4\) and \(y=-1\).

\(t=-1:\ (-2, 2)\); \(t=0:\ (0, 1)\); \(t=2:\ (4, -1)\)

- What coordinates should the line have when the parameter is zero? - The direction vector tells you how each coordinate changes with the parameter.

1. Start at the given point and multiply each component of the direction vector by \(t\). 2. The equations are \(x=-2+3t\) and \(y=1-t\), with \(t\in\mathbb{R}\).

\(x=-2+3t,\ y=1-t\), where \(t\in\mathbb{R}\)

- Look for the coordinate equation that is easiest to solve for \(t\). - Replace that parameter in the other coordinate equation.

1. Solve \(x=1+t\) for \(t\): \(t=x-1\). 2. Substitute into the other equation: \(y=2+3(x-1)=3x-1\).

\(y=3x-1\)

- 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 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 \(g\).

- 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 \(\vec{x}=t\begin{pmatrix}0\\1\end{pmatrix}\).

- Find the change in each coordinate from \(P\) to \(Q\). - Use one given point as the value reached when the parameter is zero. - For part b, choose a parameter value that does not reproduce either given endpoint.

1. A direction vector from \(P\) to \(Q\) is \((-4, 8)\). 2. One parametrization is \(x=7-4t,\ y=-3+8t\). 3. Choose a parameter value other than \(0\) or \(1\), such as \(t=2\). Then \(S=(-1, 13)\).

a) One answer is \(x=7-4t,\ y=-3+8t\), where \(t\in\mathbb{R}\). b) One possible point is \(S=(-1, 13)\).

- Build the parametrization from the displacement \(B-A\). - Test the candidate point in each coordinate equation separately. - Membership requires the same parameter value in both coordinates.

1. A direction vector from \(A\) to \(B\) is \((3, -3)\), so one parametrization is \(x=1+3t\), \(y=4-3t\). 2. For \(C=(-2, 7)\), the x-coordinate gives \(-2=1+3t\), so \(t=-1\). 3. The y-coordinate gives \(7=4-3t\), which also gives \(t=-1\). Because one common parameter value satisfies both coordinates, \(C\) lies on \(h\).

Yes. The common parameter value is \(t=-1\), so \(C=(-2, 7)\) lies on \(h\).

- A point and a direction vector determine the two coordinate equations directly. - In a parametrization based at \(P\), compare the displacement \(t\vec{v}\) with each given displacement vector. - After identifying the parameter, interpret its sign before substituting it into the coordinate equations.

1. One parametrization is \(x=2+4t,\ y=-1-3t\). 2. In this parametrization, the displacement from \(P\) is \(t\vec{v}\). Thus \(Q\) corresponds to \(t=\frac{5}{2}>0\), so it lies in the direction of \(\vec{v}\). Substitution gives \(Q=(12, -\frac{17}{2})\). 3. Point \(R\) corresponds to \(t=-2<0\), so it lies in the direction opposite \(\vec{v}\). Substitution gives \(R=(-6, 5)\).

a) \(x=2+4t,\ y=-1-3t\), where \(t\in\mathbb{R}\) b) \(Q\): \(t=\frac{5}{2}>0\), so the displacement is in the direction of \(\vec{v}\); \(Q=(12, -\frac{17}{2})\). \(R\): \(t=-2<0\), so the displacement is opposite the direction of \(\vec{v}\); \(R=(-6, 5)\).

- 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) \(g_3:\vec{x}=\begin{pmatrix}7\\0\end{pmatrix}+t\begin{pmatrix}0\\1\end{pmatrix}\)

- A point on a parametrized line must come from one common parameter value in both coordinate equations. - After finding a parameter value, compare it with the stated interval. - A point can lie on the full line without lying on the restricted traced portion.

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

a) \(P\) and \(Q\) lie on \(g\); \(R\) does not. b) Only \(P\) lies on the portion traced when \(0\le t\le3\).

- 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 \(y=4+2(-2)=0\). 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 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 \(\mathbf{x}=\langle 5,-2,1.2\rangle+t\langle 3,4,-0.1\rangle\), where \(t\ge0\). 2. Use the known \(x\)-coordinate: \(5+3t=17\), so \(t=4\). 3. Then \(y=-2+4\cdot4=14\) and \(z=1.2+4(-0.1)=0.8\). 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 \(\lambda\in\mathbb{R}\)

- 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 \(t\in\mathbb{R}\)

- 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 \(\mathbf{x}=\langle2,0,3\rangle+k\langle1,1,1\rangle+m\langle-1,2,0\rangle\). 2. For \(k=4\) and \(m=-2\), \(\mathbf{p}=\langle2,0,3\rangle+4\langle1,1,1\rangle-2\langle-1,2,0\rangle=\langle8,0,7\rangle\). 3. For \(k=\frac32\) and \(m=\frac32\), \(\mathbf{p}=\langle2,0,3\rangle+\frac32\langle1,1,1\rangle+\frac32\langle-1,2,0\rangle=\left\langle2,\frac92,\frac92\right\rangle\).

a) \(\mathbf{p}=\langle8,0,7\rangle\) b) \(\mathbf{p}=\left\langle2,\frac92,\frac92\right\rangle\)

- 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.

- First isolate \(t\) in the simpler coordinate equation. - After eliminating \(t\), do not discard the original restriction on the parameter. - Translate the two endpoint values of \(t\) into endpoint values of \(x\).

1. From \(x=t+1\), solve for the parameter: \(t=x-1\). 2. Substitute into the equation for \(y\): \(y=(x-1)^2-4\). 3. Because \(x=t+1\) increases with \(t\), the endpoints \(t=-2\) and \(t=3\) give \(x=-1\) and \(x=4\).

a) \(y=(x-1)^2-4\) b) \(-1\le x\le4\)

- Evaluate both coordinate equations at the left endpoint of the parameter interval. - To decide orientation, consider the sign of \(y\) for small positive values of the parameter. - For elimination, look for a trigonometric identity after combining the two coordinate equations.

1. At \(t=0\), \(x=3\) and \(y=0\), so the starting point is \((3, 0)\). 2. For small positive \(t\), \(y=3\sin t>0\), so the motion leaves \((3, 0)\) upward and traces the circle counterclockwise. 3. Squaring and adding gives \(x^2+y^2=9(\cos^2 t+\sin^2 t)=9\).

a) \((3, 0)\) b) Counterclockwise c) \(x^2+y^2=9\)

- Use the same value of \(t\) in both coordinate equations for each row. - Compare how the x- and y-coordinates change as the table moves to larger values of \(t\).

1. At \(t=-1\), \(x=1+2(-1)=-1\) and \(y=4-(-1)=5\). 2. At \(t=0\), \(x=1\) and \(y=4\). 3. At \(t=2\), \(x=5\) and \(y=2\). 4. As \(t\) increases, \(x\) increases while \(y\) decreases, so the point moves from upper left toward lower right along the line.

The points are \((-1, 5)\), \((1, 4)\), and \((5, 2)\). As \(t\) increases, the point moves from upper left toward lower right.

- Start from the standard cosine-sine parametrization of a circle and then shift it to the required center. - Check what point your equations produce when \(t=0\). - The sign of the sine term determines whether the motion initially goes upward or downward from the rightmost point.

1. A circle of radius \(3\) centered at \((2, -1)\) can be written as \(x=2+3\cos t\), \(y=-1+3\sin t\). 2. At \(t=0\), these equations give \((5, -1)\), the required starting point. 3. For small positive \(t\), the y-coordinate increases while the point leaves the rightmost point of the circle, so the motion is counterclockwise. 4. The interval \(0\le t\le2\pi\) completes one full revolution.

\(x=2+3\cos t\), \(y=-1+3\sin t\), for \(0\le t\le2\pi\)

- A small table with columns for \(t\), \(x\), and \(y\) can help keep the points in order. - Look at how both coordinates change between consecutive points.

1. At \(t=-1\), the point is \((0,-2)\); at \(t=0\), it is \((1,0)\); at \(t=1\), it is \((2,2)\). 2. These points lie on one straight line. The parameter traces the segment from \((0,-2)\) to \((2,2)\), moving right and up.

Ordered points: \((0,-2)\), \((1,0)\), \((2,2)\). Segment from \((0,-2)\) to \((2,2)\), moving right and up.

- Find both coordinates for each listed value of \(t\). - Plot the ordered points mentally or on paper; decide whether they fit a line or a curve. - Pay attention to the lowest point as the parameter increases.

1. Substitution gives \((-2,3)\), \((-1,0)\), \((0,-1)\), \((1,0)\), and \((2,3)\) in order. 2. Since \(x=t\), the equation is \(y=x^2-1\) on \(-2\le x\le2\), an upward-opening parabola. 3. The path moves down to the vertex \((0,-1)\), then upward as \(t\) increases.

\((-2,3)\), \((-1,0)\), \((0,-1)\), \((1,0)\), \((2,3)\); upward-opening parabolic arc, turning upward at \((0,-1)\).

- Evaluate both coordinate pairs at the first and last allowed parameter values. - Compare the endpoints before deciding whether the paths are different curves or different ways to travel one segment.

1. Path \(A\) starts at \((0,1)\) and ends at \((2,5)\). 2. Path \(B\) starts at \((2,5)\) and ends at \((0,1)\). 3. Both have the same straight-line points between these endpoints, but they are traced in opposite directions.

Yes. \(A\) goes from \((0,1)\) to \((2,5)\); \(B\) goes from \((2,5)\) to \((0,1)\). Their directions are opposite.

- Check the range of x-coordinates and y-coordinates in the arc. - Test a start value, a middle value, and an end value of each candidate parameter interval. - The direction from right to left matters as well as the shape.

1. The arc has horizontal radius \(2\), vertical radius \(1\), and stays above the x-axis. 2. At \(t=0\), choice C starts at \((2,0)\). At \(t=\frac{\pi}{2}\), it reaches \((0,1)\), and at \(t=\pi\), it ends at \((-2,0)\). 3. A traces a different half, B is below the axis, and D traces the entire ellipse. Thus C is correct.

C. \(x=2\cos t,\ y=\sin t,\ 0\le t\le\pi\); start \((2,0)\), end \((-2,0)\).

- How could you describe both paths using only their x- and y-coordinates? - To compare arrival times, find the parameter each robot has when it reaches the target point.

1. On Atlas's path, \(t=x\), so \(y=x^2\) with \(0\le x\le2\). 2. On Pixel's path, \(s=\frac{x}{2}\), so \(y=4(\frac{x}{2})^2=x^2\) on the same interval. The paths match. 3. Atlas reaches \((1,1)\) at \(t=1\) second. Pixel reaches it at \(s=\frac{1}{2}\) second, so Pixel arrives first.

Yes, both follow \(y=x^2\) for \(0\le x\le2\). Pixel reaches \((1,1)\) first, at \(\frac{1}{2}\,\text{s}\); Atlas takes \(1\,\text{s}\).

- Cartesian coordinates come from projecting the radius onto the horizontal and vertical axes. - How can you use a single parameter to let the angle vary through the stated interval?

1. Use \(x=r\cos\theta\) and \(y=r\sin\theta\). 2. Let \(t=\theta\). Then \(x(t)=(2+\cos t)\cos t\) and \(y(t)=(2+\cos t)\sin t\). 3. The full curve uses \(0\le t\le2\pi\).

\(x(t)=(2+\cos t)\cos t,\ y(t)=(2+\cos t)\sin t,\ 0\le t\le2\pi\)

- Replace each of the three coordinates in the solid's inequalities with its expression in \(t\). - Look for parameter values that satisfy all restrictions at once. - The endpoints of that shared interval determine the two boundary points.

1. The x-restriction gives \(-1\le t\le3\), the y-restriction gives \(0\le t\le3\), and the z-restriction gives \(1\le t\le2\). 2. Their overlap is \(1\le t\le2\). 3. At \(t=1\) the line enters at \((2,1,2)\); at \(t=2\) it leaves at \((3,2,0)\).

\(1\le t\le2\); entry \((2,1,2)\), exit \((3,2,0)\)

- Read each endpoint as an ordered pair from the graph. - How much does each coordinate change between the start and end of the segment? - Use the midpoint parameter to check your equations.

1. The endpoints on the graph are \(S=(-1,-2)\) and \(T=(3,0)\). 2. From \(S\) to \(T\), the x-coordinate increases by \(4\) and the y-coordinate increases by \(2\). Thus \(x(t)=-1+4t\) and \(y(t)=-2+2t\) for \(0\le t\le1\). 3. At \(t=\frac{1}{2}\), the point is \((1,-1)\).

\(x(t)=-1+4t\), \(y(t)=-2+2t\), \(0\le t\le1\); at \(t=\frac{1}{2}\): \((1,-1)\).

- Which given vertex lets you find the unknown coefficient in the plane equation? - Check the proposed plane rule at the other two vertices. - At the intersection, the moving point must satisfy the plane equation as well as the line equations.

1. Substituting \(S=(2,2,4)\) into \(z=kx\) gives \(4=2k\), so \(k=2\). 2. Both \(A\) and \(D\) have \(x=z=0\), so they also satisfy \(z=2x\). 3. On the line, \(z=2t\) and \(x=3-t\). The plane equation becomes \(2t=2(3-t)\), giving \(t=\frac{3}{2}\). 4. Substitution into the line gives \(\left(\frac{3}{2},2,3\right)\).

a) \(k=2\), so the plane is \(z=2x\); all three vertices satisfy this equation. b) \(t=\frac{3}{2}\), intersection \(\left(\frac{3}{2},2,3\right)\).

- 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 \((4, 3, 5)\).

- 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 \(S\).

- 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 \(y=z\).

- 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) \(D=(5, 4, 2)\)

- 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 \(E:\vec{x}=\begin{pmatrix}2\\1\\3\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\-1\end{pmatrix}+\sigma\begin{pmatrix}0\\2\\1\end{pmatrix}\).

- Compare the direction vectors first. - Test whether a point from one line lies on the other line. - How many planes can contain a single line? - For part b, choose a plane normal perpendicular to the line's direction vector.

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so the lines are parallel or identical. 2. The point \((3, 1, 1)\) on \(h\) lies on \(g\) when \(r=1\). Therefore, the lines are identical. 3. One line lies in infinitely many planes, so the identical lines do not uniquely determine a plane. 4. A normal vector for a plane containing the line must be perpendicular to \(\langle2,-1,1\rangle\). Choose \(\mathbf{n}=\langle1,2,0\rangle\), since their dot product is \(0\). 5. Using the point \((1, 2, 0)\), the plane equation is \(x+2y=5\).

a) The lines are identical, so they do not uniquely determine a plane. b) One possible plane is \(E:x+2y=5\).

- 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 \((7, 0, 3)\).

- 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.

- 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) \(k=15\)

- Compare both the point each parametrization starts from and the direction in which its parameter moves. - Substitute the coordinates of \(Q\) into each pair of equations. - For part c, compare the displacement from \((3, 1)\) produced by each parameter.

1. Both parametrizations use the point \((3, 1)\). Their direction vectors are \((1, 2)\) and \((-2, -4)\), and \((-2, -4)=-2(1, 2)\). Therefore, they describe the same line. 2. In \(g\), \(3+r=5\), so \(r=2\). In \(h\), \(3-2s=5\), so \(s=-1\). 3. For the same point, the parameter displacements satisfy \(r(1, 2)=s(-2, -4)\). Thus \(r=-2s\), or \(s=-\frac{1}{2}r\). The negative sign means increasing the two parameters moves in opposite directions along the line.

a) They have the same base point, and their direction vectors are nonzero scalar multiples. b) \(r=2\) and \(s=-1\) c) \(r=-2s\), or \(s=-\frac{1}{2}r\); the negative sign indicates opposite orientations.

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

- A correct parametrization needs both a point on the intended line and a direction parallel to that line. - Find a direction vector from \(A\) to \(B\) before checking either proposal. - Check whether Tom’s base point has a recognizable relationship to \(A\) and \(B\).

1. A direction vector from \(A\) to \(B\) is \((-2, 4)\). 2. Mia uses base point \((3, 0)=A\). Her direction vector is \((-1, 2)=\frac{1}{2}(-2, 4)\), so her parametrization is correct. 3. Tom uses base point \((2, 2)\), the midpoint of \(A\) and \(B\), so that point lies on the line. His direction vector is \((1, -2)=-\frac{1}{2}(-2, 4)\), so his parametrization is also correct.

Both parametrizations are correct. Mia uses point \(A\) and a parallel direction vector; Tom uses the midpoint \((2, 2)\) and the opposite parallel direction.

- 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.

- Compute \(B-A\) before substituting any candidate point. - One parameter value must satisfy both coordinate equations for a point on the containing line. - The interval \(0\le t\le1\) is what distinguishes the segment from the rest of the line.

1. \(B-A=(-6, 6)\), so \(A+t(B-A)=(5-6t, -1+6t)\). The segment is traced by \(0\le t\le1\). 2. For \(P\), both coordinates give \(t=\frac{2}{3}\). Since \(\frac{2}{3}\in[0,1]\), \(P\) lies on the segment. 3. For \(Q\), both coordinates give \(t=-\frac{1}{3}\). Thus \(Q\) lies on the containing line but not on the segment. 4. For \(R\), both coordinates give \(t=\frac{1}{3}\). Since \(\frac{1}{3}\in[0,1]\), \(R\) lies on the segment.

a) \(P\): \(t=\frac{2}{3}\); on \(\overline{AB}\) b) \(Q\): \(t=-\frac{1}{3}\); not on \(\overline{AB}\) c) \(R\): \(t=\frac{1}{3}\); on \(\overline{AB}\)

- 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.

- For parallel lines, compare their direction vectors before considering their base points. - Use the coordinate of the direction vector that does not contain \(k\) to determine the scale factor. - After finding \(k\), decide whether the two parallel lines share a point.

1. Parallel direction vectors must be scalar multiples. Since \(-9=-3(3)\), the scale factor from the direction of \(g_k\) to the direction of \(h\) is \(-3\). 2. Therefore, \(12=-3k\), so \(k=-4\). 3. For \(k=-4\), test the point \((1, 2)\) from \(g_k\) in \(h\). The x-coordinate gives \(1=4-9s\), so \(s=\frac{1}{3}\). 4. That value gives \(y=-1+12(\frac{1}{3})=3\), not \(2\). The lines are therefore distinct and parallel.

a) \(k=-4\) b) The lines are distinct and parallel.

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

- 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.

- 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 \(\langle0,0,1\rangle\), which is not a scalar multiple of \(\langle1,1,1\rangle\). 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) \(h:\vec{x}=\begin{pmatrix}0\\3\\4\end{pmatrix}+t\begin{pmatrix}0\\-1\\1\end{pmatrix}\)

- What could go wrong if two of the intersection points were actually the same point? - Find the three pairwise intersections before deciding whether a triangle exists. - The parameter values at the two vertices on each line determine the interval for that side.

1. Equating \(g\) and \(h\) gives \(\lambda=2\) and \(\mu=0\), so their intersection is \((5, 0)\). 2. Equating \(h\) and \(k\) gives \(\mu=2\) and \(\nu=0\), so their intersection is \((3, 4)\). 3. Equating \(k\) and \(g\) gives \(\nu=-2\) and \(\lambda=0\), so their intersection is \((1, 2)\). 4. The three pairwise intersection points are distinct, so the lines form a triangle. 5. The side on \(g\) corresponds to \(0\le\lambda\le2\), the side on \(h\) to \(0\le\mu\le2\), and the side on \(k\) to \(-2\le\nu\le0\).

The lines form a triangle with vertices \((5, 0)\), \((3, 4)\), and \((1, 2)\). 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.

- Reflect one convenient point of the line through \(Q\). - Think about what point reflection does to a displacement vector between two points on the line. - For the general explanation, write a typical point as a base point plus a parameter times a direction vector.

1. The point reached at \(t=0\) is \((-1, 5)\). Reflecting it through \(Q=(1, 2)\) gives \((3, -1)\). 2. Point reflection reverses every displacement vector, so the image direction can be written as \((-2, 1)\). An equivalent parametrization may use the parallel vector \((2, -1)\). 3. One parametrization is therefore \(x=3+2s,\ y=-1-s\). 4. The two lines have parallel direction vectors. Since \(Q\) is not on \(g\), the reflection does not map \(g\) onto itself, so the lines are distinct and parallel. 5. In general, a point \(\mathbf{a}+t\mathbf{v}\) reflects through \(\mathbf{q}\) to \((2\mathbf{q}-\mathbf{a})-t\mathbf{v}\). The image direction is \(-\mathbf{v}\), which is parallel to \(\mathbf{v}\).

a) One answer is \(x=3+2s,\ y=-1-s\), where \(s\in\mathbb{R}\). b) The lines are distinct and parallel. c) Point reflection changes a direction vector \(\mathbf{v}\) to \(-\mathbf{v}\), so the image line is parallel to the original line.

- 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 \((1, 1, 1)\).

- 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) \(h:\vec{x}=\begin{pmatrix}0\\0\\5\end{pmatrix}+s\begin{pmatrix}-3\\3\\-3\end{pmatrix}\)

- When do two direction vectors generate only one direction? - Compare corresponding components to find the scale factor. - Replace the two dependent parameters with one new parameter. - Test the point in all three coordinate equations.

1. The point set describes a line when its two direction vectors are linearly dependent. Require \(\langle m,4,-2\rangle=k\langle3,-2,1\rangle\). 2. The second and third coordinates give \(k=-2\). Therefore, \(m=3(-2)=-6\). 3. For \(m=-6\), the second direction vector is \(-2\) times the first, so the two parameters combine into one. A simplified equation is \(g:\mathbf{x}=\langle4,0,-2\rangle+t\langle3,-2,1\rangle\). 4. For \(P\), the coordinate equations are \(10=4+3t\), \(-4=-2t\), and \(0=-2+t\). Each gives \(t=2\), so \(P\) lies on the line.

a) \(m=-6\) b) \(g:\mathbf{x}=\langle4,0,-2\rangle+t\langle3,-2,1\rangle\) c) Yes. Point \(P\) lies on the line when \(t=2\).

- 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 \(E\).

- 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) \((0, 0, 5)\)

- 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 \(P\).

- 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.

- 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.

- Express the tetrahedron as coordinate inequalities. - Substitute the line coordinates into each inequality. - Intersect the resulting parameter intervals. - Evaluate the line at the interval endpoints.

1. The tetrahedron consists of points satisfying \(x\ge0\), \(y\ge0\), \(z\ge0\), and \(x+y+z\le6\). 2. On the line, \(x=1+t\), \(y=1+t\), and \(z=-1+2t\). 3. The conditions \(x\ge0\) and \(y\ge0\) require \(t\ge-1\). The condition \(z\ge0\) requires \(t\ge\frac12\). 4. The final inequality gives \((1+t)+(1+t)+(-1+2t)\le6\), so \(1+4t\le6\) and \(t\le\frac54\). 5. Combining the restrictions gives \(\frac12\le t\le\frac54\). 6. At \(t=\frac12\), the point is \(P_1\left(\frac32,\frac32,0\right)\). 7. At \(t=\frac54\), the point is \(P_2\left(\frac94,\frac94,\frac32\right)\).

The line enters and exits at \(P_1\left(\frac32,\frac32,0\right)\) and \(P_2\left(\frac94,\frac94,\frac32\right)\). The portion inside corresponds to \(t\in\left[\frac12,\frac54\right]\).

- The endpoints of the trace come from the endpoints of the parameter interval. - Because one coordinate equals the parameter directly, elimination can be done by substitution. - A point can satisfy the Cartesian equation without being reached by the restricted parameter interval.

1. At \(t=-2\), the point is \((-2, 3)\). At \(t=1\), the point is \((1, 0)\). 2. Since \(x=t\), substitute \(t=x\) into \(y=t^2-1\) to obtain \(y=x^2-1\). 3. For \(Q\), the x-coordinate would require \(t=\frac{3}{2}\), which lies outside \([-2,1]\). Therefore, \(Q\) is on the full Cartesian parabola but not on the traced parametric portion.

a) \((-2, 3)\) and \((1, 0)\) b) \(y=x^2-1\) c) No. It would require \(t=\frac{3}{2}\), which is outside the allowed interval.

- Crossing the x-axis imposes a condition on the y-coordinate. - After finding a possible time, use the other coordinate equation to locate the particle. - To test the origin, both coordinate equations must hold for the same parameter value.

1. On the x-axis, \(y=0\). Solve \(t^2-4t+3=(t-1)(t-3)=0\), giving \(t=1\) and \(t=3\). 2. At \(t=1\), \(x=-2\), so the position is \((-2, 0)\). At \(t=3\), \(x=2\), so the position is \((2, 0)\). 3. The origin would require \(x=0\), which gives \(t=2\). At \(t=2\), \(y=-1\), so the particle does not pass through the origin.

a) \(t=1\) at \((-2, 0)\), and \(t=3\) at \((2, 0)\) b) No

- Eliminate the parameter enough to identify the geometric curve traced by each pair. - Look at what happens just after \(t=0\) to determine direction. - Compare the total angular change in each parametrization over the whole interval.

1. For both parametrizations, \(x^2+y^2=1\), so each stays on the unit circle and each reaches every point of that circle. 2. For \(A\), the point starts at \((1, 0)\) and moves upward for small positive \(t\), so it travels counterclockwise. 3. For \(B\), the point also starts at \((1, 0)\), but its y-coordinate is negative for small positive \(t\), so it travels clockwise. 4. In \(A\), the angle changes by \(2\pi\), so the circle is traced once. In \(B\), the effective angle changes by \(4\pi\), so the circle is traced twice.

a) Yes. Both trace the unit circle. b) \(A\) is counterclockwise; \(B\) is clockwise. c) \(A\) traces the circle once; \(B\) traces it twice.

- Look for powers of \(t\) that can be made equal without solving directly for \(t\). - A Cartesian relation can contain points that a restricted parameter interval never reaches. - For each point with \(x=4\), determine which parameter value would produce its sign of \(y\).

1. From \(x=t^2\) and \(y=t^3\), squaring the second equation gives \(y^2=t^6\), while cubing the first gives \(x^3=t^6\). Therefore, \(y^2=x^3\). 2. If \(x=4\), then \(t^2=4\), so \(t=-2\) or \(t=2\). 3. The interval \([-2,1]\) allows \(t=-2\) but not \(t=2\). At \(t=-2\), \(y=-8\). 4. Therefore, \((4, -8)\) is traced, while \((4, 8)\) lies on the full Cartesian curve but is not traced by the restricted parametrization.

a) \(y^2=x^3\) b) Only \((4, -8)\) is traced. The point \((4, 8)\) would require \(t=2\), which is outside \([-2,1]\).

- Test the parameter value at the start of the interval against the labeled starting point. - The sign of \(y\) distinguishes the upper semicircle from the lower semicircle. - Check whether the stated interval stops after half a revolution or continues around the full circle.

1. The graph shows that the path starts at \((-3, 0)\). At \(t=0\), choices B, C, and D start there. 2. The shown path stays on the upper semicircle, so \(y\ge0\). Choice C has \(y=-3\sin t\le0\) on \([0,\pi]\), so it traces the lower semicircle. 3. Choice D continues through a full \(2\pi\) parameter interval and traces the entire circle, not only the shown upper half. 4. Choice B starts at \((-3, 0)\), keeps \(y\ge0\), and ends at \((3, 0)\) when \(t=\pi\). Therefore, B matches the path.

B. \(x=-3\cos t\), \(y=3\sin t\), for \(0\le t\le\pi\)

- Translate the geometric condition “lies on the line” into an equation involving the two coordinate functions. - Solve the resulting equation for the parameter, then use the stated time interval to discard any invalid value. - Substitute the valid time into both coordinate equations to locate the particle.

1. On the line \(y=x+1\), the coordinate functions must satisfy \(2t-3=t^2-4t+1\). 2. Rearranging gives \(t^2-6t+4=0\), so \(t=3\pm\sqrt{5}\). 3. Only \(t=3-\sqrt{5}\) lies in \([0,4]\). 4. At this time, \(x=2-2\sqrt{5}\) and \(y=3-2\sqrt{5}\).

\(t=3-\sqrt{5}\), at position \((2-2\sqrt{5}, 3-2\sqrt{5})\)

- At the crossing, what value must the cart's vertical coordinate have? - Compare its position at the start and at the end of the given time interval. - How can you find the length of a straight path from its horizontal and vertical changes?

1. The cart crosses \(y=4\) when \(10-3t=4\), so \(t=2\). Then \(x=4+2(2)=8\), giving the point \((8,4)\). 2. At \(t=0\) the cart is at \((4,10)\). At \(t=4\) it is at \((12,-2)\). 3. Since its path is a straight segment, the distance traveled is \(\sqrt{(12-4)^2+(-2-10)^2}=\sqrt{208}=4\sqrt{13}\) feet.

a) After \(2\,\text{s}\), at \((8,4)\) feet. b) Position \((12,-2)\) feet; distance traveled \(4\sqrt{13}\,\text{ft}\).

- Identify the Cartesian equation of the circle from its center and radius. - Replace the Cartesian coordinate expressions with their polar equivalents. - Which angles keep the resulting radius nonnegative?

1. Squaring and adding the translated coordinates gives \((x-2)^2+y^2=4\). 2. Expanding yields \(x^2+y^2=4x\). With \(x=r\cos\theta\) and \(x^2+y^2=r^2\), this becomes \(r^2=4r\cos\theta\). 3. For nonzero points, \(r=4\cos\theta\). With \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\), the radius is nonnegative; the endpoints both give the pole. This traces the whole circle once.

\(r=4\cos\theta\), with \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\), including the pole.

- Start with the three coordinate planes that form the sides at the origin. - What equation do all three nonzero vertices satisfy? - Replace the three coordinates in the solid's inequalities by their expressions in \(t\). - The ends of the allowed parameter interval give the entry and exit points.

1. The solid lies in the first octant, so \(x\ge0\), \(y\ge0\), and \(z\ge0\). Its slanted face is \(x+y+z=4\), and the solid is on the origin side. 2. Thus \(x+y+z\le4\) is the fourth inequality. 3. Substitute the line coordinates. The conditions become \(t\ge0\), \(1\ge0\), \(1\ge0\), and \(t+2\le4\). Thus \(0\le t\le2\). 4. At \(t=0\), the point is \((0,1,1)\), and at \(t=2\), it is \((2,1,1)\).

a) \(x\ge0,\ y\ge0,\ z\ge0,\ x+y+z\le4\) b) \(0\le t\le2\); entry \((0,1,1)\), exit \((2,1,1)\).

- Form two different direction vectors along the face, both starting from \(A\). - How could you check that a proposed normal vector is perpendicular to each direction? - For the edge, what changes in each coordinate as you move from \(A\) to \(C\)? - Test the plane equation for a general value of \(t\), not just one endpoint.

1. The two face directions are \(\overrightarrow{AB}=(0,3,0)\) and \(\overrightarrow{AC}=(2,1,5)\). 2. If \(\mathbf{n}=(a,b,c)\) is perpendicular to both, then \(3b=0\) and \(2a+b+5c=0\). Choose \(\mathbf{n}=(-5,0,2)\). 3. Through \(A=(1,0,0)\), the plane equation is \(-5(x-1)+2z=0\), or \(5x-2z=5\). 4. The edge from \(A\) to \(C\) has \(x=1+2t\), \(y=t\), and \(z=5t\), for \(0\le t\le1\). 5. Substitution gives \(5(1+2t)-2(5t)=5\) for every \(t\), so the whole edge lies in the plane.

a) One normal is \(\mathbf{n}=(-5,0,2)\); plane \(5x-2z=5\). b) \(x=1+2t\), \(y=t\), \(z=5t\), for \(0\le t\le1\). The plane equation reduces to \(5=5\) for all these values.

- 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.

- Solve each pair of line equations to find a common point. - Form side vectors from the vertices. - Use the dot-product angle formula after the vertices are known. - The centroid is the coordinate average of the three vertices. - A parallel line can use a scalar multiple of the side vector.

1. Pairwise solution of the coordinate equations gives \(g_1\cap g_2=B(4,2,2)\), \(g_2\cap g_3=C(1,5,2)\), and \(g_3\cap g_1=A(1,2,-1)\). The three points are distinct, so the lines form a triangle. 2. The side vectors are \(\overrightarrow{AB}=\langle 3,0,3\rangle\), \(\overrightarrow{AC}=\langle 0,3,3\rangle\), and \(\overrightarrow{BC}=\langle -3,3,0\rangle\). 3. At \(A\), \(\cos A=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{\|\overrightarrow{AB}\|\|\overrightarrow{AC}\|}=\frac{9}{\sqrt{18}\sqrt{18}}=\frac12\), so \(A=60^\circ\). 4. The same calculation at \(B\) and \(C\) gives \(B=C=60^\circ\). Thus, the triangle is equilateral. 5. The centroid is the coordinate average: \(S=\frac13(A+B+C)=(2,3,1)\). 6. A line through \(S\) parallel to \(AB\) may use \(\langle 1,0,1\rangle\) as its direction: \(h:\mathbf{r}(u)=\langle 2,3,1\rangle+u\langle 1,0,1\rangle\).

a) The lines intersect pairwise in three distinct points. b) \(A(1,2,-1)\), \(B(4,2,2)\), \(C(1,5,2)\) c) \(A=B=C=60^\circ\); the triangle is equilateral. d) \(S(2,3,1)\) e) \(h:\mathbf{r}(u)=\langle 2,3,1\rangle+u\langle 1,0,1\rangle\)

- Identify the two side faces crossed by a line parallel to the x-axis. - Find a normal vector to each face using two directions in the face. - Intersect the line with each face. - Compute the distance between the two boundary points.

1. The line crosses the side faces \(ADS\) and \(BCS\). 2. For face \(ADS\), directions are \(\overrightarrow{AD}=\langle 0,4,0\rangle\) and \(\overrightarrow{AS}=\langle 2,2,8\rangle\). A normal vector orthogonal to both is \(\langle -4,0,1\rangle\), giving the plane equation \(-4x+z=-8\). 3. On \(h\), \(x=s\) and \(z=2\). Thus, \(-4s+2=-8\), so \(s=\frac52\). The first boundary point is \(P_1\left(\frac52,4,2\right)\). 4. For face \(BCS\), a normal vector is \(\langle 4,0,1\rangle\), giving \(4x+z=24\). 5. Substitution gives \(4s+2=24\), so \(s=\frac{11}{2}\). The second boundary point is \(P_2\left(\frac{11}{2},4,2\right)\). 6. The distance between the points is \(\frac{11}{2}-\frac52=3\).

\(3\) units

- Read the center and the two semiaxis lengths from the ellipse equation. - A cosine coordinate naturally begins at a horizontal extreme when the parameter is zero. - Compare the sign of the sine term with the direction the point should initially move from the rightmost point.

1. The ellipse has center \((2, -1)\), horizontal semiaxis \(3\), and vertical semiaxis \(2\). 2. One parametrization starting at the rightmost point is \(x=2+3\cos t\) and \(y=-1-2\sin t\). 3. Using \(0\le t\le2\pi\) traces the ellipse once. At \(t=0\), the point is \((5, -1)\). For small positive \(t\), the y-coordinate decreases, so the motion begins downward from the rightmost point and is clockwise.

One valid answer is \(x=2+3\cos t\), \(y=-1-2\sin t\), for \(0\le t\le2\pi\).

- Think of the right semicircle as half of a radius-\(5\) circle with an angle that changes through exactly \(\pi\) radians. - Choose the starting angle to produce the lower endpoint and the ending angle to produce the upper endpoint. - Check the sign of \(x\) over the entire parameter interval, not just at the endpoints.

1. Let the polar angle increase from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) as \(t\) increases from \(0\) to \(1\). One choice is \(\theta=-\frac{\pi}{2}+\pi t\). 2. Using \(x=5\cos\theta\) and \(y=5\sin\theta\) gives \(x=5\sin(\pi t)\) and \(y=-5\cos(\pi t)\). 3. For \(0\le t\le1\), \(\sin(\pi t)\ge0\), so \(x\ge0\) and only the right semicircle is traced. 4. At \(t=0\), the point is \((0, -5)\); at \(t=1\), it is \((0, 5)\). The angle increases continuously by \(\pi\), so the semicircle is traced exactly once in the required direction.

One valid answer is \(x=5\sin(\pi t)\), \(y=-5\cos(\pi t)\), for \(0\le t\le1\).