Chapter 10.3, Problem 18E

To determine

To find: The product $u\times v$ if $u=\frac{2}{3}i$ and $v=\frac{1}{3}j-3k$ . Show that the product is orthogonal to both $u$ and $v$ .

Answer to Problem 18E

The product $u\times v$ is $-2j+\frac{2}{9}k$ or $\langle 0,-2,\frac{2}{9}\rangle$ .

Explanation of Solution

Given information:

The given vectors are $u=\frac{2}{3}i$ and $v=\frac{1}{3}j-3k$ .

Calculation:

Calculate the cross product of $u$ and $v$ .

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ \frac{2}{3}& 0& 0\\ 0& \frac{1}{3}& -3\end{array}\right|\\ =\left|\begin{array}{cc}0& 0\\ \frac{1}{3}& -3\end{array}\right|i-\left|\begin{array}{cc}\frac{2}{3}& 0\\ 0& -3\end{array}\right|j+\left|\begin{array}{cc}\frac{2}{3}& 0\\ 0& \frac{1}{3}\end{array}\right|k\\ =\left(0\right)i-\left(\frac{2}{3}\cdot \left(-3\right)\right)j+\left(\frac{2}{3}\cdot \frac{1}{3}\right)k\\ =2j+\frac{2}{9}k\text{or}\langle 0,2,\frac{2}{9}\rangle \end{array}$

So, the cross product $u\times v$ is $2j+\frac{2}{9}k$ .

To check the cross product is orthogonal to the vectors $u$ and $v$ , find the dot product of each vector with the cross product.

  $\begin{array}{c}u\cdot \left(u\times v\right)=\left(\frac{2}{3}i\right)\cdot \left(0i+2j+\frac{2}{9}k\right)\\ =0\end{array}$

and,

  $\begin{array}{c}v\cdot \left(u\times v\right)=\left(\frac{1}{3}j-3k\right)\cdot \left(2j+\frac{2}{9}k\right)\\ =\left(\frac{1}{3}\right)\left(2\right)-3\left(\frac{2}{9}\right)\\ =\frac{2}{3}-\frac{2}{3}\\ =0\end{array}$

From the above, it is clear that the product of each vector with the cross product is zero. Hence, proved.