# 2.4 The cross product  (Page 3/16)

 Page 3 / 16

For property $\text{iv}.,$ this follows directly from the definition of the cross product. We have

$\begin{array}{cc}\hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0& =⟨{u}_{2}\left(0\right)-{u}_{3}\left(0\right),\text{−}\left({u}_{2}\left(0\right)-{u}_{3}\left(0\right)\right),{u}_{1}\left(0\right)-{u}_{2}\left(0\right)⟩\hfill \\ & =⟨0,0,0⟩=0.\hfill \end{array}$

Then, by property i., $0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}=0$ as well. Remember that the dot product of a vector and the zero vector is the scalar $0,$ whereas the cross product of a vector with the zero vector is the vector $0.$

Property $\text{vi}.$ looks like the associative property, but note the change in operations:

$\begin{array}{cc}\hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =\text{u}·⟨{v}_{2}{w}_{3}-{v}_{3}{w}_{2},\text{−}{v}_{1}{w}_{3}+{v}_{3}{w}_{1},{v}_{1}{w}_{2}-{v}_{2}{w}_{1}⟩\hfill \\ & ={u}_{1}\left({v}_{2}{w}_{3}-{v}_{3}{w}_{2}\right)+{u}_{2}\left(\text{−}{v}_{1}{w}_{3}+{v}_{3}{w}_{1}\right)+{u}_{3}\left({v}_{1}{w}_{2}-{v}_{2}{w}_{1}\right)\hfill \\ & ={u}_{1}{v}_{2}{w}_{3}-{u}_{1}{v}_{3}{w}_{2}-{u}_{2}{v}_{1}{w}_{3}+{u}_{2}{v}_{3}{w}_{1}+{u}_{3}{v}_{1}{w}_{2}-{u}_{3}{v}_{2}{w}_{1}\hfill \\ & =\left({u}_{2}{v}_{3}-{u}_{3}{v}_{2}\right){w}_{1}+\left({u}_{3}{v}_{1}-{u}_{1}{v}_{3}\right){w}_{2}+\left({u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right){w}_{3}\hfill \\ & =⟨{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}⟩·⟨{w}_{1},{w}_{2},{w}_{3}⟩\hfill \\ & =\left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)·\text{w}.\hfill \end{array}$

## Using the properties of the cross product

Use the cross product properties to calculate $\left(2\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}.$

$\begin{array}{cc}\hfill \left(2\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}& =2\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\hfill \\ & =2\left(3\right)\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\hfill \\ & =\left(6\text{k}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\hfill \\ & =6\left(\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right)\hfill \\ & =6\left(\text{−}\text{i}\right)=-6\text{i}.\hfill \end{array}$

Use the properties of the cross product to calculate $\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right).$

$\text{−}\text{k}$

So far in this section, we have been concerned with the direction of the vector $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v},$ but we have not discussed its magnitude. It turns out there is a simple expression for the magnitude of $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ involving the magnitudes of $\text{u}$ and $\text{v},$ and the sine of the angle between them.

## Magnitude of the cross product

Let $\text{u}$ and $\text{v}$ be vectors, and let $\theta$ be the angle between them. Then, $‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖=‖\text{u}‖·‖\text{v}‖·\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$

## Proof

Let $\text{u}=⟨{u}_{1},{u}_{2},{u}_{3}⟩$ and $\text{v}=⟨{v}_{1},{v}_{2},{v}_{3}⟩$ be vectors, and let $\theta$ denote the angle between them. Then

$\begin{array}{cc}\hfill {‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖}^{2}& ={\left({u}_{2}{v}_{3}-{u}_{3}{v}_{2}\right)}^{2}+{\left({u}_{3}{v}_{1}-{u}_{1}{v}_{3}\right)}^{2}+{\left({u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right)}^{2}\hfill \\ & ={u}_{2}^{2}{v}_{3}^{2}-2{u}_{2}{u}_{3}{v}_{2}{v}_{3}+{u}_{3}^{2}{v}_{2}^{2}+{u}_{3}^{2}{v}_{1}^{2}-2{u}_{1}{u}_{3}{v}_{1}{v}_{3}+{u}_{1}^{2}{v}_{3}^{2}+{u}_{1}^{2}{v}_{2}^{2}-2{u}_{1}{u}_{2}{v}_{1}{v}_{2}+{u}_{2}^{2}{v}_{1}^{2}\hfill \\ & ={u}_{1}^{2}{v}_{1}^{2}+{u}_{1}^{2}{v}_{2}^{2}+{u}_{1}^{2}{v}_{3}^{2}+{u}_{2}^{2}{v}_{1}^{2}+{u}_{2}^{2}{v}_{2}^{2}+{u}_{2}^{2}{v}_{3}^{2}+{u}_{3}^{2}{v}_{1}^{2}+{u}_{3}^{2}{v}_{2}^{2}+{u}_{3}^{2}{v}_{3}^{2}\hfill \\ & \phantom{\rule{2em}{0ex}}-\left({u}_{1}^{2}{v}_{1}^{2}+{u}_{2}^{2}{v}_{2}^{2}+{u}_{3}^{2}{v}_{3}^{2}+2{u}_{1}{u}_{2}{v}_{1}{v}_{2}+2{u}_{1}{u}_{3}{v}_{1}{v}_{3}+2{u}_{2}{u}_{3}{v}_{2}{v}_{3}\right)\hfill \\ & =\left({u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}\right)\left({v}_{1}^{2}+{v}_{2}^{2}+{v}_{3}^{2}\right)-{\left({u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3}\right)}^{2}\hfill \\ & ={‖\text{u}‖}^{2}{‖\text{v}‖}^{2}-{\left(\text{u}·\text{v}\right)}^{2}\hfill \\ & ={‖\text{u}‖}^{2}{‖\text{v}‖}^{2}-{‖\text{u}‖}^{2}{‖\text{v}‖}^{2}{\text{cos}}^{2}\theta \hfill \\ & ={‖\text{u}‖}^{2}{‖\text{v}‖}^{2}\left(1-{\text{cos}}^{2}\theta \right)\hfill \\ & ={‖\text{u}‖}^{2}{‖\text{v}‖}^{2}\left({\text{sin}}^{2}\theta \right).\hfill \end{array}$

Taking square roots and noting that $\sqrt{{\text{sin}}^{2}\theta }=\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ for $0\le \theta \le 180\text{°},$ we have the desired result:

$‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖=‖\text{u}‖‖\text{v}‖\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$

This definition of the cross product allows us to visualize or interpret the product geometrically. It is clear, for example, that the cross product is defined only for vectors in three dimensions, not for vectors in two dimensions. In two dimensions, it is impossible to generate a vector simultaneously orthogonal to two nonparallel vectors.

## Calculating the cross product

Use [link] to find the magnitude of the cross product of $\text{u}=⟨0,4,0⟩$ and $\text{v}=⟨0,0,-3⟩.$

We have

$\begin{array}{cc}\hfill ‖\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}‖& =‖\text{u}‖·‖\text{v}‖·\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \hfill \\ & =\sqrt{{0}^{2}+{4}^{2}+{0}^{2}}·\sqrt{{0}^{2}+{0}^{2}+{\left(-3\right)}^{2}}·\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi }{2}\hfill \\ & =4\left(3\right)\left(1\right)=12.\hfill \end{array}$

Use [link] to find the magnitude of $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v},$ where $\text{u}=⟨-8,0,0⟩$ and $\text{v}=⟨0,2,0⟩.$

$16$

## Determinants and the cross product

Using [link] to find the cross product of two vectors is straightforward, and it presents the cross product in the useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have an alternative. We can calculate the cross product of two vectors using determinant    notation.

A $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinant is defined by

$|\begin{array}{cc}{a}_{1}\hfill & {a}_{2}\hfill \\ {b}_{1}\hfill & {b}_{2}\hfill \end{array}|={a}_{1}{b}_{2}-{b}_{1}{a}_{2}.$

For example,

$|\begin{array}{cc}3\hfill & \hfill -2\\ 5\hfill & \hfill 1\end{array}|=3\left(1\right)-5\left(-2\right)=3+10=13.$

A $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant is defined in terms of $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinants as follows:

$|\begin{array}{ccc}{a}_{1}\hfill & {a}_{2}\hfill & {a}_{3}\hfill \\ {b}_{1}\hfill & {b}_{2}\hfill & {b}_{3}\hfill \\ {c}_{1}\hfill & {c}_{2}\hfill & {c}_{3}\hfill \end{array}|={a}_{1}|\begin{array}{cc}{b}_{2}\hfill & {b}_{3}\hfill \\ {c}_{2}\hfill & {c}_{3}\hfill \end{array}|-{a}_{2}|\begin{array}{cc}{b}_{1}\hfill & {b}_{3}\hfill \\ {c}_{1}\hfill & {c}_{3}\hfill \end{array}|+{a}_{3}|\begin{array}{cc}{b}_{1}\hfill & {b}_{2}\hfill \\ {c}_{1}\hfill & {c}_{2}\hfill \end{array}|.$

[link] is referred to as the expansion of the determinant along the first row . Notice that the multipliers of each of the $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinants on the right side of this expression are the entries in the first row of the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant. Furthermore, each of the $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinants contains the entries from the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant that would remain if you crossed out the row and column containing the multiplier. Thus, for the first term on the right, ${a}_{1}$ is the multiplier, and the $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinant contains the entries that remain if you cross out the first row and first column of the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant. Similarly, for the second term, the multiplier is ${a}_{2},$ and the $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2$ determinant contains the entries that remain if you cross out the first row and second column of the $3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3$ determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.

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