Free math worksheets

- 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 \((3, 5)\).

\((3, 5)\)

- Form the displacement vector from \(A\) to \(B\) first. - Collinearity here is equivalent to one displacement vector being a scalar multiple of another. - A single scalar must work in every component.

1. \(\mathbf{v}=\overrightarrow{AB}=\langle2,-4,4\rangle\). 2. \(\overrightarrow{AC}=\langle-1,2,-2\rangle=-\frac12\mathbf{v}\), so \(C\) is collinear with \(A\) and \(B\). 3. \(\overrightarrow{AD}=\langle3,-6,7\rangle\). The first two components would require the scalar \(\frac32\), but \(\frac32\cdot4=6\ne7\). Therefore, \(\overrightarrow{AD}\) is not a scalar multiple of \(\mathbf{v}\), so \(D\) is not collinear with \(A\) and \(B\).

a) Yes; \(\overrightarrow{AC}=-\frac12\mathbf{v}\). b) No; \(\overrightarrow{AD}\) is not a scalar multiple of \(\mathbf{v}\).

- Compute the reference displacement vector once and compare every component with it. - In part a, two known components determine the only possible scalar before the missing coordinate is found. - Opposite directions correspond to a negative scalar.

1. \(\overrightarrow{PQ}=\langle3,3,-3\rangle\). 2. \(\overrightarrow{PR}=\langle a-2,6,-6\rangle\). The second and third components show that the scalar must be \(2\). Therefore, \(a-2=2\cdot3=6\), so \(a=8\). 3. \(\overrightarrow{PS}=\langle-3,-3,3\rangle=-\overrightarrow{PQ}\). Thus, \(\overrightarrow{PS}\) is a scalar multiple of \(\overrightarrow{PQ}\).

a) \(a=8\), with scalar \(2\) b) Yes; \(\overrightarrow{PS}=-\overrightarrow{PQ}\).

- Convert the stated horizontal and vertical changes into a slope. - Use the given point to determine the constant term in \(y=mx+b\). - At the x-intercept, the y-coordinate is zero.

1. The slope is \(m=\frac{-6}{3}=-2\). 2. Substitute \((-2, 5)\) into \(y=-2x+b\): \(5=-2(-2)+b\), so \(b=1\). Thus, \(f(x)=-2x+1\). 3. At the x-intercept, \(0=-2x+1\), so \(x=\frac{1}{2}\). The intercept is \(\left(\frac{1}{2}, 0\right)\).

a) \(f(x)=-2x+1\) b) \(\left(\frac{1}{2}, 0\right)\)

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

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

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

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

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

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

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

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

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

- Read each coordinate condition as a restriction on vector components. - A direction parallel to a coordinate axis has only the corresponding component nonzero. - A direction parallel to a horizontal plane has zero vertical component.

1. Along the x-axis, a valid choice is \(P=(0, 0, 0)\) and \(\mathbf{u}=\langle1,0,0\rangle\). 2. Parallel to the z-axis, use the required point \(P=(2, 5, 0)\) and a vector with only a z-component, such as \(\mathbf{u}=\langle0,0,1\rangle\). 3. To stay in the xz-plane, both y-components must be zero. One valid choice is \(P=(1, 0, 0)\) and \(\mathbf{u}=\langle1,0,1\rangle\). 4. For a direction parallel to \(z=4\), the z-component of the direction must be zero. One valid choice is \(P=(0, 0, 4)\) and \(\mathbf{u}=\langle1,1,0\rangle\).

Answers vary. One set of valid choices is: a) \(P=(0, 0, 0)\), \(\mathbf{u}=\langle1,0,0\rangle\) b) \(P=(2, 5, 0)\), \(\mathbf{u}=\langle0,0,1\rangle\) c) \(P=(1, 0, 0)\), \(\mathbf{u}=\langle1,0,1\rangle\) d) \(P=(0, 0, 4)\), \(\mathbf{u}=\langle1,1,0\rangle\)

- 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 \((-2, 0)\).

- Magnitude comes from the square root of the sum of the squared components. - Scaling a vector by a positive scalar preserves its direction and scales its magnitude by the same factor. - A unit vector has magnitude \(1\).

1. \(\|\mathbf{v}\|=\sqrt{2^2+4^2+(-6)^2}=\sqrt{56}=2\sqrt{14}\). 2. A vector in the same direction with half the magnitude is \(\frac12\mathbf{v}=\langle 1,2,-3\rangle\), whose magnitude is \(\sqrt{14}\). 3. Divide \(\mathbf{v}\) by its magnitude: \(\frac{\mathbf{v}}{\|\mathbf{v}\|}=\left\langle\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}\right\rangle\).

a) \(2\sqrt{14}\) b) \(\langle 1,2,-3\rangle\) c) \(\left\langle\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}\right\rangle\)

- 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 \(g\).

- Equating the three coordinates produces three linear equations in the two line parameters. - Keep the augmented column visible while row-reducing; consistency is the key question. - If the system is consistent, substitute the solved parameters into either original line to recover the point.

1. Equating coordinates gives \(2r+s=3\), \(r-2s=-1\), and \(-3r+4s=1\). 2. Row-reduce \(\left[\begin{array}{cc|c}2&1&3\\1&-2&-1\\-3&4&1\end{array}\right]\). Its reduced form is \(\left[\begin{array}{cc|c}1&0&1\\0&1&1\\0&0&0\end{array}\right]\). 3. Thus \(r=1\) and \(s=1\). Substitution into either line gives \((3, 2, 1)\).

The lines intersect at \((3, 2, 1)\), with \(r=1\) and \(s=1\).

- Treat the line parameters as the two unknowns of a linear system. - A contradiction row after row reduction proves that no common parameter pair exists. - After establishing nonintersection, compare the direction vectors before deciding between parallel and skew.

1. Equating coordinates gives \(\lambda-3\mu=2\), \(2\lambda+\mu=-3\), and \(2\lambda-\mu=4\). 2. Row-reducing \(\left[\begin{array}{cc|c}1&-3&2\\2&1&-3\\2&-1&4\end{array}\right]\) produces a contradiction row of the form \(\left[\begin{array}{cc|c}0&0&1\end{array}\right]\), so the system is inconsistent. 3. The direction vectors \(\langle 1,2,2\rangle\) and \(\langle 3,-1,1\rangle\) are not scalar multiples, so the lines are not parallel. Nonparallel lines in space that do not intersect are skew.

The augmented system is inconsistent, so the lines do not intersect. Their direction vectors are not parallel, so the lines are skew.

- Compute each displacement vector before comparing directions. - Dependence of two nonzero vectors means that one is a scalar multiple of the other. - A single mismatched zero/nonzero component can rule out a scalar multiple immediately.

1. In part a, \(\overrightarrow{AB}=\langle1,2,1\rangle\) and \(\overrightarrow{AC}=\langle3,6,3\rangle=3\overrightarrow{AB}\). The directions are dependent. 2. In part b, \(\overrightarrow{GP}=\langle-1,1,0\rangle\). This cannot be a scalar multiple of \(\langle0,1,1\rangle\) because its first component is nonzero while every scalar multiple of \(\mathbf{u}\) has first component zero. The directions are independent.

a) Dependent; \(\overrightarrow{AC}=3\overrightarrow{AB}\). b) Independent; no scalar multiple of \(\langle0,1,1\rangle\) equals \(\langle-1,1,0\rangle\).

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

- Form the displacement vector from the starting point to the sensor. - One scalar must multiply every component of \(\mathbf{u}\) to obtain that displacement. - Interpret the sign of the scalar geometrically after finding it.

1. \(\overrightarrow{LS}=\langle-1-2,7-1,8-5\rangle=\langle-3,6,3\rangle\). 2. \(\langle-3,6,3\rangle=3\langle-1,2,1\rangle=3\mathbf{u}\). 3. The scalar is \(3\), which is nonnegative, so the displacement is allowed. A negative scalar would point in the direction opposite \(\mathbf{u}\).

Yes. \(\overrightarrow{LS}=3\mathbf{u}\), and \(3\ge0\), so the displacement is in the allowed direction.

- A displacement vector is found by subtracting initial coordinates from terminal coordinates. - Translation by a vector means adding that vector to the position vector of the point. - For the final check, compare two displacement vectors rather than using a line equation.

1. \(\overrightarrow{AB}=B-A=\langle 3,2,2\rangle\). 2. \(\overrightarrow{OP}=\langle 6,4,4\rangle=2\langle 3,2,2\rangle=2\overrightarrow{AB}\). 3. \(\overrightarrow{OP'}=\overrightarrow{OP}+\overrightarrow{OA}=\langle 6,4,4\rangle+\langle 2,-1,5\rangle=\langle 8,3,9\rangle\), so \(P'=(8,3,9)\). 4. \(\overrightarrow{AP'}=P'-A=\langle 6,4,4\rangle=\overrightarrow{OP}\).

a) \(\langle 3,2,2\rangle\) b) \(\overrightarrow{OP}=2\overrightarrow{AB}\) c) \(P'=(8,3,9)\) d) \(\overrightarrow{AP'}=\langle 6,4,4\rangle=\overrightarrow{OP}\)

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

- A vector in the xz-plane must have a zero y-component. - Rewrite the linear relation as a ratio between the x- and z-components. - In part b, the same scalar must work for all three components.

1. Because \(\mathbf{u}\) lies in the xz-plane, \(u_y=0\). The relation \(5u_x=2u_z\) is satisfied by the smallest positive integer pair \(u_x=2\), \(u_z=5\). Thus, one direction vector is \(\langle2,0,5\rangle\). 2. \(\langle4,0,10\rangle=2\langle2,0,5\rangle\), so \(\mathbf{q}\) is a scalar multiple with scalar \(2\).

a) \(\langle2,0,5\rangle\) b) Yes; \(\mathbf{q}=2\mathbf{u}\).

- 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) \(\mathbf{p}=\left\langle2,\frac92,\frac92\right\rangle\)

- Form both displacement vectors from the same initial point. - The first two components determine the only possible scalar multiple. - Use the final component to determine the unknown coordinate, then interpret the scalar-multiple relationship geometrically.

1. \(\overrightarrow{AB}=\langle2,1,2\rangle\) and \(\overrightarrow{AC_k}=\langle4,2,k-1\rangle\). 2. The first two components require \(\overrightarrow{AC_k}=2\overrightarrow{AB}\). 3. Therefore, \(k-1=2\cdot2=4\), so \(k=5\). 4. When \(k=5\), \(\overrightarrow{AC_5}=2\overrightarrow{AB}\), so \(A\), \(B\), and \(C_5\) are collinear.

\(k=5\). The scalar is \(2\), and the three points are collinear.

- 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.

- A position reached by a translation is found by adding the translation vector to the starting position vector. - The fourth vertex must include both side translations from \(A\). - For part b, subtract the initial endpoint's position vector from the terminal endpoint's position vector. - Compare the two three-vector sums component by component in part c.

1. \(\overrightarrow{OB}=\mathbf{a}+\mathbf{u}=\langle5,1,0\rangle\) and \(\overrightarrow{OD}=\mathbf{a}+\mathbf{v}=\langle2,2,-2\rangle\). 2. The fourth vertex is \(\overrightarrow{OC}=\mathbf{a}+\mathbf{u}+\mathbf{v}=\langle4,3,0\rangle\). 3. \(\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\langle-1,2,0\rangle=\mathbf{v}\), and \(\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=\langle2,1,2\rangle=\mathbf{u}\). Thus opposite sides have equal direction vectors. 4. Since vector addition is commutative, \(\mathbf{a}+\mathbf{u}+\mathbf{v}=\mathbf{a}+\mathbf{v}+\mathbf{u}=\langle4,3,0\rangle\). The two translations reach the same final point in either order.

a) \(\overrightarrow{OB}=\langle5,1,0\rangle\), \(\overrightarrow{OD}=\langle2,2,-2\rangle\), \(\overrightarrow{OC}=\langle4,3,0\rangle\) b) \(\overrightarrow{BC}=\langle-1,2,0\rangle=\mathbf{v}\) and \(\overrightarrow{DC}=\langle2,1,2\rangle=\mathbf{u}\); opposite sides have matching direction vectors. c) Both sums equal \(\langle4,3,0\rangle\), so the translations may be performed in either order.

- Opposite sides of a parallelogram have the same displacement vector. - From \(B\), use the displacement that takes \(A\) to \(D\). - From \(D\), use the displacement that takes \(A\) to \(B\). - For the final part, subtract position vectors in terminal-minus-initial order.

1. Since \(\overrightarrow{AD}=3\mathbf{v}\), moving from \(B\) by the same vector gives \(\overrightarrow{OC}=\overrightarrow{OB}+3\mathbf{v}=\mathbf{a}+2\mathbf{u}+3\mathbf{v}\). 2. Since \(\overrightarrow{AB}=2\mathbf{u}\), moving from \(D\) by the same vector gives \(\overrightarrow{OC}=\overrightarrow{OD}+2\mathbf{u}=\mathbf{a}+3\mathbf{v}+2\mathbf{u}\). Commutativity of vector addition makes this equal to \(\mathbf{a}+2\mathbf{u}+3\mathbf{v}\). 3. \(\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=3\mathbf{v}\), and \(\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=2\mathbf{u}\).

a) \(\overrightarrow{OC}=\mathbf{a}+2\mathbf{u}+3\mathbf{v}\) b) \(\overrightarrow{OC}=\mathbf{a}+3\mathbf{v}+2\mathbf{u}=\mathbf{a}+2\mathbf{u}+3\mathbf{v}\) c) \(\overrightarrow{BC}=3\mathbf{v}\), \(\overrightarrow{DC}=2\mathbf{u}\)

- 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 \(0\le t\le2\pi\)

- Equating the three coordinates produces three linear equations in \(r\), \(s\), and \(k\). - Keep the variable order \(r,s,k\) fixed when you build the augmented matrix. - Use row reduction to solve the complete system before substituting into either line.

1. Equating coordinates gives \(r-2s=2\), \(r-s=2\), and \(2r-k=0\). 2. The augmented matrix is \(\left[\begin{array}{ccc|c}1&-2&0&2\\1&-1&0&2\\2&0&-1&0\end{array}\right]\), with columns for \(r,s,k\). 3. Row reduction gives \(\left[\begin{array}{ccc|c}1&0&0&2\\0&1&0&0\\0&0&1&4\end{array}\right]\). Therefore, \(r=2\), \(s=0\), and \(k=4\). 4. Substituting \(r=2\) into \(g\) gives \(S=(4, 3, 4)\).

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

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

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

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

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

1. The direction vector is \(\overrightarrow{AB}=\langle3,6,-3\rangle\). 2. For \(\lambda=0.4\), \(\mathbf{x}=\langle1,-2,3\rangle+0.4\langle3,6,-3\rangle=\langle2.2,0.4,1.8\rangle\). Thus, \(P=(2.2, 0.4, 1.8)\). 3. A point lies strictly between \(A\) and \(B\) when \(0<\lambda<1\). Its distance from \(A\) is \(\lambda\|\overrightarrow{AB}\|\), and its distance from \(B\) is \((1-\lambda)\|\overrightarrow{AB}\|\). The first is greater when \(\lambda>\frac12\). Therefore, \(\frac12<\lambda<1\). 4. When \(\lambda<0\), the point lies outside segment \(\overline{AB}\), beyond \(A\) on the side opposite \(B\).

a) \(P=(2.2, 0.4, 1.8)\) b) \(\lambda\in\left(\frac12,1\right)\) c) Point \(P\) lies beyond \(A\) on the extension of the line opposite \(B\).

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

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

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

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

1. The plane reaches the ground when its \(z\)-coordinate is \(0\). Solve \(600-20t=0\), which gives \(t=30\). Therefore, the plane reaches the ground after \(30\,\text{s}\). 2. At \(t=30\), the position is \(\mathbf{x}(30)=\langle -2000+30\cdot80,1500+30(-60),600+30(-20)\rangle=\langle 400,-300,0\rangle\). 3. The distance from the origin is \(\sqrt{400^2+(-300)^2}=\sqrt{250000}=500\).

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

- Write a general vector and impose a zero dot product with each spanning direction. - Any nonzero scalar multiple of a valid normal vector is also valid. - After finding the normal, the displacement from the known plane point to \(P\) must be orthogonal to it.

1. Let \(\mathbf{n}=\langle x,y,z\rangle\). Orthogonality gives \(x+2z=0\) and \(2x+y-z=0\). 2. One nonzero solution is \(\mathbf{n}=\langle-2,5,1\rangle\). 3. \(\overrightarrow{P_0P}=\langle3,1,a-3\rangle\). Since this displacement lies in the plane, \(\langle-2,5,1\rangle\cdot\langle3,1,a-3\rangle=0\). 4. Thus, \(-6+5+a-3=0\), so \(a=4\).

One normal vector is \(\mathbf{n}=\langle-2,5,1\rangle\), and \(a=4\).

- Translate point membership into one linear equation for each coordinate. - Put the coefficients in a consistent column order for \(a\), \(\lambda\), and \(\mu\). - Row-reduce the augmented matrix rather than solving the equations by successive substitution.

1. Equating coordinates gives \(a-\lambda=0\), \(a+\lambda-2\mu=2\), and \(2\lambda+\mu=8\). 2. With variables ordered as \(a,\lambda,\mu\), the augmented matrix is \(\left[\begin{array}{ccc|c}1&-1&0&0\\1&1&-2&2\\0&2&1&8\end{array}\right]\). 3. Row reduction gives \(\left[\begin{array}{ccc|c}1&0&0&3\\0&1&0&3\\0&0&1&2\end{array}\right]\). Thus \(a=3\), \(\lambda=3\), and \(\mu=2\). 4. These values satisfy all three coordinate equations, so the point lies in the plane when \(a=3\).

\(a=3\)

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

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

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

- The sought normal must have zero dot product with each independent tabletop direction. - Use the known support point only after the normal vector has been determined. - For the center height, substitute the given horizontal coordinates into the Cartesian equation.

1. \(\overrightarrow{P_1P_2}=\langle40,0,1\rangle\) and \(\overrightarrow{P_1P_3}=\langle0,40,-1\rangle\). 2. For \(\mathbf{n}=\langle a,b,c\rangle\), the conditions are \(40a+c=0\) and \(40b-c=0\). One solution is \(\mathbf{n}=\langle-1,1,40\rangle\). 3. A point-normal equation is \((\mathbf{x}-\langle0,0,30\rangle)\cdot\langle-1,1,40\rangle=0\). Expanding gives \(-x+y+40z=1200\). 4. At \(x=20\) and \(y=20\), \(-20+20+40z=1200\), so \(z=30\).

a) \(\overrightarrow{P_1P_2}=\langle40,0,1\rangle\), \(\overrightarrow{P_1P_3}=\langle0,40,-1\rangle\), and one normal is \(\mathbf{n}=\langle-1,1,40\rangle\). b) \((\mathbf{x}-\langle0,0,30\rangle)\cdot\langle-1,1,40\rangle=0\); equivalently, \(-x+y+40z=1200\) c) \(30\,\text{in.}\)

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

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

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

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

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

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

- The axis intercepts occur where two coordinates are zero; the given trace parametrizations make those values accessible directly. - Use two independent displacement vectors lying in the plane before looking for a normal. - A plane normal has zero dot product with every direction lying in that plane.

1. The common x-axis intercept is \(S_1=(5, 0, 0)\). Setting \(r=1\) gives \(S_2=(0, 2, 0)\), and setting \(s=1\) gives \(S_3=(0, 0, 10)\). 2. \(\mathbf{u}=\langle-5,2,0\rangle\) and \(\mathbf{v}=\langle-5,0,10\rangle\). For \(\mathbf{n}=\langle a,b,c\rangle\), orthogonality gives \(-5a+2b=0\) and \(-5a+10c=0\). 3. One nonzero solution is \(\mathbf{n}=\langle2,5,1\rangle\). 4. Using \(S_1\), \((\mathbf{x}-\langle5,0,0\rangle)\cdot\langle2,5,1\rangle=0\), which expands to \(2x+5y+z=10\). 5. For \(Q\), \(2(1)+5(1)+3=10\), so \(Q\) lies in the plane.

a) \(S_1=(5, 0, 0)\), \(S_2=(0, 2, 0)\), \(S_3=(0, 0, 10)\) b) One normal vector is \(\mathbf{n}=\langle2,5,1\rangle\). c) \(2x+5y+z=10\); yes, \(Q\) lies in the plane.

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

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

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

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

1. Let \(\mathbf{u}=\langle2,1,0\rangle\) and \(\mathbf{v}=\langle-1,2,0\rangle\). Their dot product is \(\mathbf{u}\cdot\mathbf{v}=2(-1)+1\cdot2+0=0\), so they are perpendicular. 2. Their lengths are equal: \(\|\mathbf{u}\|=\sqrt5\) and \(\|\mathbf{v}\|=\sqrt5\). Both parameters vary across intervals of length \(2\), so the four sides are equal and adjacent sides are perpendicular. Thus, \(M\) is a square. 3. The parameter intervals are centered at \(r=0\) and \(s=0\), so the center is \((0, 0, 5)\). 4. Each side has length \(2\sqrt5\). Therefore, the area is \((2\sqrt5)^2=20\) square units.

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

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

1. Equating corresponding coordinates gives \(r=2s\), \(1+2r=6-s\), and \(-1+2r=2+s\). Solving gives \(s=1\) and \(r=2\), so \(S=(4, 5, 3)\). 2. The direction vectors \(\langle1,2,2\rangle\) and \(\langle2,-1,1\rangle\) are not scalar multiples. Since the lines intersect and have different directions, they determine a unique plane. 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be a normal vector. Perpendicularity to both line directions gives \(a+2b+2c=0\) and \(2a-b+c=0\). One solution is \(\mathbf{n}=\langle4,3,-5\rangle\). Using a point on the plane gives \(E:4x+3y-5z=16\).

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

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

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so the lines are parallel. 2. The point \((1, 0, 3)\) is not on \(h\): the first coordinate would require \(k=\frac12\), while the second would require \(k=-1\). Thus, the lines are distinct parallel lines. 3. Use \(\mathbf{u}=\langle2,-1,4\rangle\) and the vector between the given points, \(\mathbf{v}=\langle2,2,-2\rangle\), as spanning vectors. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both. Then \(2a-b+4c=0\) and \(2a+2b-2c=0\). One solution is \(\mathbf{n}=\langle-1,2,1\rangle\). 4. Using \((1, 0, 3)\), the plane equation is \(F:-x+2y+z=2\).

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

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

1. The first coordinate would require \(0=2+t\), so \(t=-2\). This value gives \(y=5-2(-2)=9\), not \(1\). Therefore, \(A\notin h\). A line and a point not on it determine a unique plane. 2. Use the line direction \(\mathbf{u}=\langle1,-2,3\rangle\) and the connecting vector \(\mathbf{v}=\langle-2,-4,5\rangle\). Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both. Then \(a-2b+3c=0\) and \(-2a-4b+5c=0\). One solution is \(\mathbf{n}=\langle2,-11,-8\rangle\). 3. Using point \(A\), the plane equation is \(E:2x-11y-8z=-43\).

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

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

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

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

- A normal to the plane must be orthogonal to both independent displacement vectors in the plane. - Turn each orthogonality statement into a dot product equal to zero. - Point membership can be checked only after the plane equation has been established.

1. \(\overrightarrow{PQ}=\langle-2,3,-1\rangle\) and \(\overrightarrow{PR}=\langle1,-1,1\rangle\). 2. For \(\mathbf{n}=\langle a,b,c\rangle\), orthogonality gives \(-2a+3b-c=0\) and \(a-b+c=0\). One solution is \(\mathbf{n}=\langle2,1,-1\rangle\). 3. A point-normal equation is \((\mathbf{x}-\langle4,0,1\rangle)\cdot\langle2,1,-1\rangle=0\). Expanding gives \(2x+y-z=7\). 4. For \(S\), \(2(1)+4-(-1)=7\), so \(S\) lies in the plane.

a) One normal vector is \(\mathbf{n}=\langle2,1,-1\rangle\). b) \((\mathbf{x}-\langle4,0,1\rangle)\cdot\langle2,1,-1\rangle=0\); equivalently, \(2x+y-z=7\) c) Yes, \(S\) lies in the plane.

- The normal must have zero dot product with each given direction. - Solve the two homogeneous equations for any convenient nonzero normal vector. - Use the shared point only after the normal has been found.

1. For \(\mathbf{n}=\langle a,b,c\rangle\), the equations \(\mathbf{n}\cdot\mathbf{u}=0\) and \(\mathbf{n}\cdot\mathbf{v}=0\) are \(2a+b=0\) and \(-b+c=0\). 2. One nonzero solution is \(\mathbf{n}=\langle1,-2,-2\rangle\). 3. A point-normal equation is \((\mathbf{x}-\langle1,0,2\rangle)\cdot\langle1,-2,-2\rangle=0\). Expanding gives \(x-2y-2z=-3\). 4. For \(T\), \(5-2(1)-2(3)=-3\), so \(T\) lies in the plane.

a) One normal vector is \(\mathbf{n}=\langle1,-2,-2\rangle\). b) \((\mathbf{x}-\langle1,0,2\rangle)\cdot\langle1,-2,-2\rangle=0\); equivalently, \(x-2y-2z=-3\) c) Yes, \(T\) lies in the plane.

- The vector from the known point in the plane to the second point gives another direction contained in the plane. - A normal vector must be perpendicular to both independent plane directions. - Express perpendicularity with two dot products equal to zero.

1. \(\mathbf{v}=\overrightarrow{GP}=\langle4,-3,0\rangle\). 2. For \(\mathbf{n}=\langle a,b,c\rangle\), orthogonality gives \(a-2b+2c=0\) and \(4a-3b=0\). One solution is \(\mathbf{n}=\langle6,8,5\rangle\). 3. Using \(G\), a point-normal equation is \((\mathbf{x}-\langle0,3,1\rangle)\cdot\langle6,8,5\rangle=0\).

a) \(\mathbf{v}=\langle4,-3,0\rangle\) b) One normal vector is \(\mathbf{n}=\langle6,8,5\rangle\). c) \((\mathbf{x}-\langle0,3,1\rangle)\cdot\langle6,8,5\rangle=0\)

- 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.

- Compute the magnitude of the base vector before considering any scalar multiple. - Write the coordinates of \(s\mathbf{d}\) explicitly for the coordinate-sum condition. - The magnitude of a scalar multiple depends on the absolute value of the scalar.

1. \(\|\mathbf{d}\|=\sqrt{3^2+(-2)^2+6^2}=\sqrt{49}=7\). 2. \(P_s=(3s,-2s,6s)\), so the coordinate sum is \(7s\). Thus, \(7s=21\), giving \(s=3\) and \(P_3=(9,-6,18)\). 3. \(\|s\mathbf{d}\|=|s|\|\mathbf{d}\|=7|s|\). Setting \(7|s|=35\) gives \(s=5\) or \(s=-5\). 4. The corresponding points are \((15,-10,30)\) and \((-15,10,-30)\).

a) \(7\) b) \(s=3\), so \(P_s=(9,-6,18)\) c) \(s=5\) or \(s=-5\), giving \((15,-10,30)\) and \((-15,10,-30)\)

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

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

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

- First convert the total displacement from \(A\) to \(B\) into displacement per second. - The position function starts at \(A\) and adds the elapsed-time multiple of the constant velocity. - Use the altitude component to determine the requested time. - For part c), distinguish the velocity vector from its scalar magnitude.

1. The displacement from \(A\) to \(B\) is \(\langle30,30,-30\rangle\). Over \(6\,\text{s}\), the constant velocity is \(\langle5,5,-5\rangle\,\text{m/s}\). 2. Therefore, \(\mathbf{r}(t)=\langle10+5t,20+5t,100-5t\rangle\) for \(0\le t\le6\). 3. Set the altitude equal to \(85\): \(100-5t=85\), so \(t=3\,\text{s}\). Then \(\mathbf{r}(3)=\langle25,35,85\rangle\), giving position \((25, 35, 85)\). 4. Differentiating componentwise gives \(\mathbf{r}'(t)=\langle5,5,-5\rangle\,\text{m/s}\). Its magnitude is \(\sqrt{5^2+5^2+(-5)^2}=5\sqrt{3}\,\text{m/s}\).

a) \(\mathbf{r}(t)=\langle10+5t,20+5t,100-5t\rangle\), \(0\le t\le6\) b) \(t=3\,\text{s}\), at \((25, 35, 85)\) c) Velocity \(\langle5,5,-5\rangle\,\text{m/s}\); speed \(5\sqrt{3}\,\text{m/s}\)

- 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.

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

- Translate the side relation of the parallelogram directly into a vector equation. - The midpoint of points with position vectors \(\mathbf{p}\) and \(\mathbf{q}\) has position vector \(\frac12(\mathbf{p}+\mathbf{q})\). - Keep point coordinates and displacement vectors distinct when subtracting.

1. \(\overrightarrow{BC}=C-B=\langle-2,4,-2\rangle\), so \(D=A+\overrightarrow{BC}=(0, 3, -2)\). 2. \(A+C=(6, 4, 2)\) and \(B+D=(6, 4, 2)\). Dividing either equal coordinate sum by \(2\) gives the same midpoint for \(\overline{AC}\) and \(\overline{BD}\), so the diagonals bisect each other. 3. \(\overrightarrow{BD}=D-B=\langle-6,2,-6\rangle\).

a) \(D=(0, 3, -2)\) b) \(A+C=B+D=(6, 4, 2)\), so both diagonal midpoints are \((3, 2, 1)\). c) \(\overrightarrow{BD}=\langle-6,2,-6\rangle\)

- Three lines form a triangle when each pair intersects at a different 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) \(h:\vec{x}=\begin{pmatrix}0\\0\\5\end{pmatrix}+s\begin{pmatrix}-3\\3\\-3\end{pmatrix}\)

- Use one known point as the position vector and two connecting vectors as spanning directions. - Find a normal vector that is perpendicular to both spanning directions. - Use the normal-vector components as the coefficients in the Cartesian equation.

1. Use \(A\) as a point and form the spanning vectors \(\overrightarrow{AB}=\langle2,0,-2\rangle\) and \(\overrightarrow{AC}=\langle0,3,-2\rangle\). Thus, \(E:\mathbf{x}=\langle1,2,3\rangle+r\langle2,0,-2\rangle+s\langle0,3,-2\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(2a-2c=0\) and \(3b-2c=0\). One solution is \(\mathbf{n}=\langle3,2,3\rangle\). 3. Using point \(A\), the Cartesian equation is \(E:3x+2y+3z=16\).

a) \(E:\mathbf{x}=\langle1,2,3\rangle+r\langle2,0,-2\rangle+s\langle0,3,-2\rangle\) b) \(E:3x+2y+3z=16\)

- Form two independent connecting vectors from one sail vertex. - Find a normal vector perpendicular to both spanning directions. - Convert the point-normal equation to Cartesian form before testing the sensor point.

1. Two spanning vectors are \(\overrightarrow{PQ}=\langle4,0,-4\rangle\) and \(\overrightarrow{PR}=\langle0,4,-4\rangle\). Thus, \(E:\mathbf{x}=\langle2,2,6\rangle+r\langle4,0,-4\rangle+s\langle0,4,-4\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(4a-4c=0\) and \(4b-4c=0\), so a simplified normal vector is \(\langle1,1,1\rangle\). A point-normal equation is \((\mathbf{x}-\langle2,2,6\rangle)\cdot\langle1,1,1\rangle=0\). 3. The corresponding Cartesian equation is \(x+y+z=10\). Since \(3+3+4=10\), the sensor point lies in the plane.

a) \(E:\mathbf{x}=\langle2,2,6\rangle+r\langle4,0,-4\rangle+s\langle0,4,-4\rangle\) b) \((\mathbf{x}-\langle2,2,6\rangle)\cdot\langle1,1,1\rangle=0\) c) Yes, \(S=(3,3,4)\) lies in \(E\).

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

- First compare the fixed coordinate on the line with the corresponding coordinate of \(P\). - Use the line direction and one vector from the line to \(P\) as plane-spanning directions. - Find a normal perpendicular to both spanning directions, then use one known point to determine the Cartesian constant.

1. Every point on \(g\) has \(y=2\), while \(P\) has \(y=4\). Therefore, \(P\notin g\). 2. Use the line direction \(\langle2,0,-1\rangle\) and the vector from the line's given point to \(P\), \(\langle4,2,-2\rangle\). Thus, \(E:\mathbf{x}=\langle1,2,3\rangle+r\langle2,0,-1\rangle+s\langle4,2,-2\rangle\). 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(2a-c=0\) and \(4a+2b-2c=0\). One solution is \(\mathbf{n}=\langle1,0,2\rangle\). Using \(P\), the Cartesian equation is \(E:x+2z=7\).

a) \(P\notin g\). b) \(E:\mathbf{x}=\langle1,2,3\rangle+r\langle2,0,-1\rangle+s\langle4,2,-2\rangle\) c) \(E:x+2z=7\)

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

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

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

- 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.

- Use one common initial point so all three displacement vectors can be compared in one linear-combination equation. - Put the coefficients of \(r\) and \(s\) into the first two matrix columns and \(\mathbf{w}\) into the augmented column. - Consistency of the matrix system determines whether the fourth displacement lies in the span of the first two.

1. \(\mathbf{u}=\langle 3,1,-2\rangle\), \(\mathbf{v}=\langle -1,3,-1\rangle\), and \(\mathbf{w}=\langle 5,5,-5\rangle\). 2. The system for \(r\mathbf{u}+s\mathbf{v}=\mathbf{w}\) is \(3r-s=5\), \(r+3s=5\), and \(2r+s=5\). 3. Row-reducing \(\left[\begin{array}{cc|c}3&-1&5\\1&3&5\\2&1&5\end{array}\right]\) gives \(\left[\begin{array}{cc|c}1&0&2\\0&1&1\\0&0&0\end{array}\right]\). 4. Thus \(r=2\) and \(s=1\), so \(\mathbf{w}=2\mathbf{u}+\mathbf{v}\). Therefore, \(D\) lies in the plane through \(A\), \(B\), and \(C\), and the four points are coplanar.

\(r=2\) and \(s=1\), so \(\mathbf{w}=2\mathbf{u}+\mathbf{v}\). The four points are coplanar.

- Set corresponding line coordinates equal to find the intersection parameters. - Use the intersection point and both line directions for the plane's parametric form. - Find a normal perpendicular to both directions, then use the intersection point for Cartesian form.

1. Equating coordinates gives \(3+t=2+2s\), \(4+2t=2+s\), and \(1-t=2+s\). Solving gives \(t=-1\) and \(s=0\), so \(S=(2,2,2)\). 2. The intersection point and the two line directions give \(E:\mathbf{x}=\langle2,2,2\rangle+t\langle1,2,-1\rangle+s\langle2,1,1\rangle\). 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both direction vectors. Then \(a+2b-c=0\) and \(2a+b+c=0\). One solution is \(\mathbf{n}=\langle1,-1,-1\rangle\). Using \(S\), the Cartesian equation is \(E:x-y-z=-2\).

1. \(S=(2,2,2)\) 2. \(E:\mathbf{x}=\langle2,2,2\rangle+t\langle1,2,-1\rangle+s\langle2,1,1\rangle\) 3. \(E:x-y-z=-2\)

- Compare the two line directions for scalar-multiple parallelism. - Use one line direction and a vector joining the two lines as plane-spanning directions. - Find a normal perpendicular to those two directions and use a known point for Cartesian form.

1. The first direction vector is \(-2\) times the second, so the lines are parallel. The point \((0,1,0)\) is not on \(g_1\): the \(x\)-coordinate would require \(r=-\frac12\), which gives \(y=3\), not \(1\). Thus, the lines are distinct. 2. Use direction \(\mathbf{u}=\langle-1,1,-2\rangle\) and the vector between the given points, \(\mathbf{v}=\langle-1,-1,-3\rangle\). One equation is \(E:\mathbf{x}=\langle0,1,0\rangle+s\langle-1,1,-2\rangle+k\langle-1,-1,-3\rangle\). 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(-a+b-2c=0\) and \(-a-b-3c=0\). One solution is \(\mathbf{n}=\langle-5,-1,2\rangle\). Using \((0,1,0)\) gives \(-5x-y+2z=-1\), or equivalently \(E:5x+y-2z=1\).

1. The lines are distinct and parallel. 2. \(E:\mathbf{x}=\langle0,1,0\rangle+s\langle-1,1,-2\rangle+k\langle-1,-1,-3\rangle\) 3. \(E:5x+y-2z=1\)

- Use the line direction and a vector from the line to the additional point as plane-spanning directions. - Find a normal perpendicular to both directions. - Use that normal for Cartesian form, then test \(P\) by substitution.

1. Use the line direction \(\mathbf{u}=\langle1,0,1\rangle\) and the vector from the line's given point to \(Q\), \(\mathbf{v}=\langle1,-3,3\rangle\). Thus, \(E:\mathbf{x}=\langle1,2,0\rangle+r\langle1,0,1\rangle+s\langle1,-3,3\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(a+c=0\) and \(a-3b+3c=0\). One solution is \(\mathbf{n}=\langle3,-2,-3\rangle\). Using \((1,2,0)\), the Cartesian equation is \(E:3x-2y-3z=-1\). 3. For \(P\), \(3\cdot4-2\cdot(-1)-3\cdot5=-1\). Therefore, \(P\in E\).

1. \(E:\mathbf{x}=\langle1,2,0\rangle+r\langle1,0,1\rangle+s\langle1,-3,3\rangle\) 2. \(E:3x-2y-3z=-1\) 3. Yes, \(P=(4,-1,5)\) lies in \(E\).

- Use one point and two connecting vectors for parametric form. - Find a normal perpendicular to both spanning vectors. - Use the Cartesian equation for the reverse membership condition on \(D\).

1. Two spanning vectors are \(\overrightarrow{AB}=\langle-2,2,-1\rangle\) and \(\overrightarrow{AC}=\langle-4,0,4\rangle\). Thus, \(E:\mathbf{x}=\langle4,1,1\rangle+r\langle-2,2,-1\rangle+s\langle-4,0,4\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(-2a+2b-c=0\) and \(-4a+4c=0\). One solution is \(\mathbf{n}=\langle2,3,2\rangle\). Using \(A\), the Cartesian equation is \(E:2x+3y+2z=13\). 3. Substitute \(D\): \(2\cdot1+3\cdot2+2z=13\). Thus, \(8+2z=13\), so \(z=\frac52\).

a) \(E:\mathbf{x}=\langle4,1,1\rangle+r\langle-2,2,-1\rangle+s\langle-4,0,4\rangle\) b) \(E:2x+3y+2z=13\) c) \(z=\frac52\)

- 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 \(t\in\left[\frac12,\frac54\right]\).

- Read the two normal vectors from the plane equations. - A direction in the intersection line is orthogonal to both normals. - Solve the two dot-product equations for a nonzero direction vector. - Find one common point by fixing one coordinate.

1. The normal vectors are \(\mathbf{n}_1=\langle 1,1,1\rangle\) and \(\mathbf{n}_2=\langle 2,-1,1\rangle\). 2. Let the line direction be \(\mathbf{d}=\langle a,b,c\rangle\). It must satisfy \(a+b+c=0\) and \(2a-b+c=0\). 3. Subtracting the equations gives \(a-2b=0\), so \(a=2b\). The first equation then gives \(c=-3b\). Choose \(b=1\), giving \(\mathbf{d}=\langle 2,1,-3\rangle\). 4. To find a point on both planes, set \(z=0\). Then \(x+y=6\) and \(2x-y=3\), so \(x=3\) and \(y=3\). 5. Therefore \(g:\mathbf{r}(t)=\langle 3,3,0\rangle+t\langle 2,1,-3\rangle\).

\(g:\mathbf{r}(t)=\langle 3,3,0\rangle+t\langle 2,1,-3\rangle\)

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

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

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

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

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

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

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

1. The base is centered at the origin and has side length \(8\), so each base coordinate is \(4\) units from the corresponding axis. One consistent labeling is \(A(4,-4,0)\), \(B(4,4,0)\), \(C(-4,4,0)\), and \(D(-4,-4,0)\). 2. A parametric equation of the line through \(S\) and \(A\) is \(\mathbf{x}=\langle 0,0,12\rangle+t\langle 4,-4,-12\rangle\). The edge corresponds to \(0\le t\le1\). 3. Substituting \(P\) gives \(2=4t\), \(-2=-4t\), and \(6=12-12t\). Each equation gives \(t=\frac12\). 4. Because \(\frac12\) is between \(0\) and \(1\), point \(P\) lies on edge \(\overline{SA}\).

a) One consistent labeling is \(A(4,-4,0)\), \(B(4,4,0)\), \(C(-4,4,0)\), and \(D(-4,-4,0)\). b) Yes. Point \(P(2,-2,6)\) lies on \(\overline{SA}\) when \(t=\frac12\).

- 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 \((2-2\sqrt{5}, 3-2\sqrt{5})\)

- 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) \(C=(1, 0, 4)\)

- Build one system in the two line parameters, but keep \(a\) in the augmented column. - Row reduction should produce one row whose consistency depends only on \(a\). - Once the system is consistent, use the solved line parameters to recover the common point.

1. Equating coordinates gives \(\lambda-3\mu=a-1\), \(\lambda+\mu=1\), and \(\lambda=-1\). 2. The augmented matrix is \(\left[\begin{array}{cc|c}1&-3&a-1\\1&1&1\\1&0&-1\end{array}\right]\). Subtract the third row from the first two to obtain rows \([0,-3\mid a]\) and \([0,1\mid2]\). 3. Adding three times the second of these rows to the first gives \([0,0\mid a+6]\). Consistency therefore requires \(a+6=0\), so \(a=-6\). 4. Then \(\lambda=-1\) and \(\mu=2\). Substitution gives the intersection point \((0, 6, 1)\).

\(a=-6\), and the lines intersect at \((0, 6, 1)\).

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

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

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

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

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

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

- 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 two nonparallel directions contained in the plane before looking for a normal. - Translate perpendicularity to both directions into two zero-dot-product equations. - Use the resulting normal with point \(P\) to write the plane equation.

1. The common line direction \(\mathbf{u}\) lies in the plane. Because the lines are distinct, \(\overrightarrow{PQ}\) supplies another direction in the plane that is not parallel to \(\mathbf{u}\). A plane normal must therefore be perpendicular to both. 2. \(\overrightarrow{PQ}=\langle-1,3,1\rangle\). For \(\mathbf{n}=\langle a,b,c\rangle\), the conditions are \(2a+b=0\) and \(-a+3b+c=0\). 3. One solution is \(\mathbf{n}=\langle1,-2,7\rangle\). 4. Using \(P\), \((\mathbf{x}-\langle1,-1,2\rangle)\cdot\langle1,-2,7\rangle=0\), which expands to \(x-2y+7z=17\).

a) The plane normal must be orthogonal to both independent directions contained in the plane. b) One normal vector is \(\mathbf{n}=\langle1,-2,7\rangle\). c) \(x-2y+7z=17\)

- 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

- A line lies in a plane when one point is in the plane and its direction is orthogonal to the plane's normal. - The second plane needs a different normal, but that normal must still be perpendicular to the same line direction. - Distinct nonparallel plane normals ensure that the two planes intersect in a line rather than coincide.

1. \(P\) satisfies \(E_1\) because \(2(2)-0=4\). Also, \(\mathbf{n}_1\cdot\mathbf{d}=\langle 2,-1,0\rangle\cdot\langle 1,2,-1\rangle=2-2=0\). Thus the direction of \(g\) is parallel to \(E_1\), and because \(P\in E_1\), the whole line lies in \(E_1\). 2. One choice is \(\mathbf{n}_2=\langle 1,0,1\rangle\), since \(\langle 1,0,1\rangle\cdot\langle 1,2,-1\rangle=0\). It is not a scalar multiple of \(\mathbf{n}_1\). 3. Using \(P\), \((\mathbf{x}-\langle 2,0,1\rangle)\cdot\langle 1,0,1\rangle=0\), so one possible equation is \(E_2:x+z=3\). 4. The planes are distinct because their normals are not parallel, so they intersect in one line. Both contain \(g\), so that intersection line is \(g\).

a) \(P\in E_1\) and \(\mathbf{n}_1\cdot\mathbf{d}=0\), so \(g\subset E_1\). b) One choice is \(\mathbf{n}_2=\langle 1,0,1\rangle\). c) One possible plane is \(E_2:x+z=3\). The distinct planes both contain \(g\), so their intersection is \(g\).

- 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 \(0\le t\le1\).