# 2.4 The cross product  (Page 6/16)

 Page 6 / 16

Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.

## Proof

The area of the base of the parallelepiped is given by $‖\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}‖.$ The height of the figure is given by $‖{\text{proj}}_{\text{v×w}}\text{u}‖.$ The volume of the parallelepiped is the product of the height and the area of the base, so we have

$\begin{array}{cc}\hfill V& =‖{\text{proj}}_{\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}}\text{u}‖‖\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}‖\hfill \\ & =|\frac{\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)}{‖\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}‖}|‖\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}‖\hfill \\ & =|\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)|.\hfill \end{array}$

## Calculating the volume of a parallelepiped

Let $\text{u}=⟨-1,-2,1⟩,\text{v}=⟨4,3,2⟩,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}=⟨0,-5,-2⟩.$ Find the volume of the parallelepiped with adjacent edges $\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}$ ( [link] ).

We have

$\begin{array}{cc}\hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =|\begin{array}{ccc}\hfill -1& \hfill -2& \hfill 1\\ \hfill 4& \hfill 3& \hfill 2\\ \hfill 0& \hfill -5& \hfill -2\end{array}|=\left(-1\right)|\begin{array}{cc}\hfill 3& \hfill 2\\ \hfill -5& \hfill -2\end{array}|+2|\begin{array}{cc}\hfill 4& \hfill 2\\ \hfill 0& \hfill -2\end{array}|+|\begin{array}{cc}\hfill 4& \hfill 3\\ \hfill 0& \hfill -5\end{array}|\hfill \\ & =\left(-1\right)\left(-6+10\right)+2\left(-8-0\right)+\left(-20-0\right)\hfill \\ & =-4-16-20\hfill \\ & =-40.\hfill \end{array}$

Thus, the volume of the parallelepiped is $|-40|=40$ units 3 .

Find the volume of the parallelepiped formed by the vectors $\text{a}=3\text{i}+4\text{j}-\text{k},$ $\text{b}=2\text{i}-\text{j}-\text{k},$ and $\text{c}=3\text{j}+\text{k}.$

$8$ units 3

## Applications of the cross product

The cross product appears in many practical applications in mathematics, physics, and engineering. Let’s examine some of these applications here, including the idea of torque, with which we began this section. Other applications show up in later chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields ( Introduction to Vector Calculus ).

## Using the triple scalar product

Use the triple scalar product to show that vectors $\text{u}=⟨2,0,5⟩,\text{v}=⟨2,2,4⟩,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}=⟨1,-1,3⟩$ are coplanar—that is, show that these vectors lie in the same plane.

Start by calculating the triple scalar product to find the volume of the parallelepiped defined by $\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}\text{:}$

$\begin{array}{cc}\hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =|\begin{array}{ccc}\hfill 2& \hfill 0& \hfill 5\\ \hfill 2& \hfill 2& \hfill 4\\ \hfill 1& \hfill -1& \hfill 3\end{array}|\hfill \\ & =\left[2\left(2\right)\left(3\right)+\left(0\right)\left(4\right)\left(1\right)+5\left(2\right)\left(-1\right)\right]-\left[5\left(2\right)\left(1\right)+\left(2\right)\left(4\right)\left(-1\right)+\left(0\right)\left(2\right)\left(3\right)\right]\hfill \\ & =2-2\hfill \\ & =0.\hfill \end{array}$

The volume of the parallelepiped is $0$ units 3 , so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.

Are the vectors $\text{a}=\text{i}+\text{j}-\text{k},$ $\text{b}=\text{i}-\text{j}+\text{k},$ and $\text{c}=\text{i}+\text{j}+\text{k}$ coplanar?

No, the triple scalar product is $-4\ne 0,$ so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.

## Finding an orthogonal vector

Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points $P=\left(9,-3,-2\right),Q=\left(1,3,0\right),$ and $R=\left(-2,5,0\right).$

The plane must contain vectors $\stackrel{\to }{PQ}$ and $\stackrel{\to }{QR}\text{:}$

$\begin{array}{c}\stackrel{\to }{PQ}=⟨1-9,3-\left(-3\right),0-\left(-2\right)⟩=⟨-8,6,2⟩\hfill \\ \stackrel{\to }{QR}=⟨-2-1,5-3,0-0⟩=⟨-3,2,0⟩.\hfill \end{array}$

The cross product $\stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{QR}$ produces a vector orthogonal to both $\stackrel{\to }{PQ}$ and $\stackrel{\to }{QR}.$ Therefore, the cross product is orthogonal to the plane that contains these two vectors:

$\begin{array}{cc}\hfill \stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{QR}& =|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill -8& \hfill 6& \hfill 2\\ \hfill -3& \hfill 2& \hfill 0\end{array}|\hfill \\ & =0\text{i}-6\text{j}-16\text{k}-\left(-18\text{k}+4\text{i}+0\text{j}\right)\hfill \\ & =-4\text{i}-6\text{j}+2\text{k}.\hfill \end{array}$

We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the cross product to real-world situations.

Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotational effect, called torque.

## Definition

Torque , $\tau$ (the Greek letter tau ), measures the tendency of a force to produce rotation about an axis of rotation. Let $\text{r}$ be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector $\text{F}$ represent the force. Then torque is equal to the cross product of $\text{r}$ and $\text{F}\text{:}$

$\tau =\text{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{F}.$

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