1.) Let \(\overrightarrow{a}=\left\langle 2,1,4\right\rangle\) and \(\overrightarrow{b}=\left\langle 3,-3,5\right\rangle\).

a.) Find both \(\overrightarrow{a}\times\overrightarrow{b}\) and \(\overrightarrow{b}\times\overrightarrow{a}\), showing that they are indeed opposite vectors.

\[\begin{align}\overrightarrow{a}\times\overrightarrow{b}&=\begin{vmatrix}i&j&k\\2&1&4\\3&-3&5\end{vmatrix}\\&=\begin{vmatrix}1&4\\-3&5\end{vmatrix}i-\begin{vmatrix}2&4\\3&5\end{vmatrix}j+\begin{vmatrix}2&1\\3&-3\end{vmatrix}k\\&=17i+2j-9k=\left\langle 17,2,-9\right\rangle\end{align}\]

\[\begin{align}\overrightarrow{b}\times\overrightarrow{a}&=\begin{vmatrix}i&j&k\\3&-3&5\\2&1&4\end{vmatrix}\\&=\begin{vmatrix}-3&5\\1&4\end{vmatrix}i-\begin{vmatrix}3&5\\2&4\end{vmatrix}j+\begin{vmatrix}3&-3\\2&1\end{vmatrix}k\\&=-17i-2j+9k=\left\langle -17,-2,9\right\rangle\end{align}\]

b.) Use your answers in part (a) to determine the area of the triangle whose vertices are \(\left(0,0,0\right)\), \(\left(2,1,4\right)\), and \(\left(3,-3,5\right)\).

This would be half the area of the parallelogram generated by our two vectors, so is half the magnitude of the cross product.

\[A=\frac{1}{2}\sqrt{17^{2}+2^{2}+9^{2}}\approx 9.67\]

2.) Find the volume of the parallelopiped determined by the vectors \(\overrightarrow{r}=\left\langle 2,1,6\right\rangle\), \(\overrightarrow{s}=\left\langle -1, -3, -2\right\rangle\), and \(\overrightarrow{t}=\left\langle 0,3,5\right\rangle\). What happens when \(\overrightarrow{t}\) is replaced by \(\overrightarrow{t^{\prime}}=\left\langle 0, -5, 2\right\rangle\)? Why does that happen?

We find this by taking the absolute value of the triple product \(\overrightarrow{r}\cdot\left(\overrightarrow{s}\times\overrightarrow{t}\right)\), which we can compute via determinant.

\[\begin{align}\begin{vmatrix}2&1&6\\-1&-3&-2\\0&3&5\end{vmatrix}&=2\cdot\begin{vmatrix}-3&-2\\3&5\end{vmatrix}-1\cdot\begin{vmatrix}-1&-2\\0&5\end{vmatrix}+6\cdot\begin{vmatrix}-1&-3\\0&3\end{vmatrix}\\&=2\cdot\left(-9\right)-1\cdot\left(-5\right)+6\cdot\left(-3\right)\\&=-18+5-18=-31\end{align}\]

Taking the absolute value of this gives us the volume \(31\).

Now we try with the altered vector

\[\begin{align}\begin{vmatrix}2&1&6\\-1&-3&-2\\0&-5&2\end{vmatrix}&=2\cdot\begin{vmatrix}-3&-2\\-5&2\end{vmatrix}-1\cdot\begin{vmatrix}-1&-2\\0&2\end{vmatrix}+6\cdot\begin{vmatrix}-1&-3\\0&-5\end{vmatrix}\\&=2\cdot\left(-16\right)-1\cdot\left(-2\right)+6\cdot\left(6\right)\\&=-32+2-30=0\end{align}\]

This volume is \(0\), because the new vector \(\overrightarrow{t^{\prime}}\) is in the same plane as \(\overrightarrow{r}\) and \(\overrightarrow{s}\).

3.) A wrench is attached to a bolt and makes an angle of \(20^{\circ}\) above the horizontal. A straight-downward force of 30 N is applied to the wrench at a point that is 15 cm away from the bolt. What is the magnitude of the torque about the bolt?

This configuration means that the angle between the wrench and the force is \(70^{\circ}\).

\[\begin{align}\left|\overrightarrow{\tau}\right|&=\left|\overrightarrow{r}\right|\left|\overrightarrow{F}\right|\sin\theta\\&=0.15\times 30\times \sin70^{\circ}\\&\approx 4.23 \text{ N-m}\end{align}\]

4.) A force represented by \(\overrightarrow{F}=\left\langle 12, -10, 4\right\rangle\) is applied to an object at point \(P=\left(3,4,-2\right)\). What is the magnitude of the torque about the point \(Q=\left(1,3,5\right)\)?

Here we calculate the torque via \(\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}\), where \(\overrightarrow{r}=\overrightarrow{QP}=\left\langle 2,1,-7\right\rangle\).

\[\begin{align}\overrightarrow{\tau}&=\begin{vmatrix}i&j&k\\2&1&-7\\12&-10&4\end{vmatrix}\\&=\begin{vmatrix}1&-7\\-10&4\end{vmatrix}i-\begin{vmatrix}2&-7\\12&4\end{vmatrix}j+\begin{vmatrix}2&1\\12&-10\end{vmatrix}k\\&=-66i-92j-32k\end{align}\]

We need the magnitude of this, which is \(\sqrt{66^{2}+92^{2}+32^{2}}\approx 117.66\text{ units}\).

5.) Let \(A=\left(1,-4,6\right)\) and \(B=\left(-3,2,4\right)\). Find the vector equation, parametric equations, and symmetric equations of the line \(\overleftrightarrow{AB}\).

Seeing that \(\overrightarrow{AB}=\left\langle -4,6,-2\right\rangle\), one possible vector equation is

\[\left\langle 1,-4,6\right\rangle + t\left\langle -4,6,-2\right\rangle\]

This leads to parametric equations

\[\begin{align}x&=1-4t\\y&=-4+6t\\z&=6-2t\end{align}\]

Finally, solving each for \(t\) and equating the expressions gives us the symmetric equations

\[-\frac{x-1}{4}=\frac{y+4}{6}=-\frac{z-6}{2}\]

6.) Find an equation of the plane through the points \(P=\left(1,2,3\right)\), \(Q=\left(2,-3,6\right)\), and \(R=\left(-3,5,9\right)\).

The vectors \(\overrightarrow{PQ}=\left\langle 1,-5,3\right\rangle\) and \(\overrightarrow{PR}=\left\langle -4,3,6\right\rangle\) lie in the plane, so we take their cross product to find a normal vector to the plane.

\[\begin{align}\overrightarrow{n}=\begin{vmatrix}i&j&k\\1&-5&3\\-4&3&6\end{vmatrix}\\&=\begin{vmatrix}-5&3\\3&6\end{vmatrix}i-\begin{vmatrix}1&3\\-4&6\end{vmatrix}j+\begin{vmatrix}1&-5\\-4&3\end{vmatrix}k\\&=-39i-18j-17k\end{align}\]

Negating the normal vector is still a normal vector, so I will use \(\overrightarrow{n}=\left\langle 39, 18, 17\right\rangle\).

An equation of the plane is \[39x+18y+17z=d\]

Plug in \(P\) to solve for \(d\):

\[39+36+51=d\implies d=126\]

Therefore an equation of the plane is \[39x+18y+17z = 126\]

7.) What is the angle between the planes \(2x-5y+z=16\) and \(-4x+z=12\)?

To find the angle between two planes, you find the angle between their normal vectors, in this case \(\left\langle 2,-5,1\right\rangle\) and \(\left\langle -4,0,1\right\rangle\). This gives

\[\theta=\cos^{-1}\left(\frac{-8+0+1}{\sqrt{30}\sqrt{17}}\right)\approx 108.06^{\circ}\]

8.) What is the distance from the point \(M=\left(12,51,-29\right)\) to the plane \(31x-11y+15z=100\)?

We first rewrite the plane as \(31x-11y+15z-100=0\). Then the distance is given by

\[d=\frac{\left|31\cdot 12 - 11\cdot 51 + 15\cdot \left(-29\right) - 100\right|}{\sqrt{31^{2}+11^{2}+15^{2}}}\approx 20.03\]