Chapter 10.3, Problem 17E

To determine

To find: The product $u\times v$ if $u=\frac{1}{6}k$ and $v=-6i-\frac{1}{3}j$ . Show that the product is orthogonal to both $u$ and $v$ .

Answer to Problem 17E

The product $u\times v$ is $\frac{1}{18}i-j+0k$ or $\langle \frac{1}{18},-1,0\rangle$ .

Explanation of Solution

Given information:

The given vectors are $u=\frac{1}{6}k$ and $v=-6i-\frac{1}{3}j$ .

Calculation:

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

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 0& 0& \frac{1}{6}\\ -6& -\frac{1}{3}& 0\end{array}\right|\\ =\left|\begin{array}{cc}0& \frac{1}{6}\\ -\frac{1}{3}& 0\end{array}\right|i-\left|\begin{array}{cc}0& \frac{1}{6}\\ -6& 0\end{array}\right|j+\left|\begin{array}{cc}0& 0\\ -6& -\frac{1}{3}\end{array}\right|k\\ =\left(0-\left(-\frac{1}{3}\cdot \frac{1}{6}\right)\right)i-\left(0-\left(-\frac{1}{6}\cdot 6\right)\right)j+\left(0-0\right)k\\ =\frac{1}{18}i-j+0k\text{or}\langle \frac{1}{18},-1,0\rangle \end{array}$

So, the cross product $u\times v$ is $\frac{1}{18}i-j+0k$ .

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

  $\begin{array}{c}u\cdot \left(u\times v\right)=\left(\frac{1}{6}k\right)\cdot \left(\frac{1}{18}i-j+0k\right)\\ =0-0+\left(\frac{1}{6}\cdot 0\right)\\ =0\end{array}$

and,

  $\begin{array}{c}v\cdot \left(u\times v\right)=\left(-6i-\frac{1}{3}j\right)\cdot \left(\frac{1}{18}i-j+0k\right)\\ =\left(-6\right)\cdot \left(\frac{1}{18}\right)+\left(-\frac{1}{3}\right)\left(-1\right)\\ =-\frac{1}{3}+\frac{1}{3}\\ =0\end{array}$

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