Chapter 10.3, Problem 9E

To determine

To find: The product $u\times v$ if $u=\langle -9,4,7\rangle$ and $v=\langle 6,3,2\rangle$ . Show that the product is orthogonal to both $u$ and $v$ .

Answer to Problem 9E

The product $u\times v$ is $-13i+60j-51k$ or $\langle -13,60,-51\rangle$ .

Explanation of Solution

Given information:

The given vectors are $u=\langle -9,4,7\rangle$ and $v=\langle 6,3,2\rangle$ .

Calculation:

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

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ -9& 4& 7\\ 6& 3& 2\end{array}\right|\\ =\left|\begin{array}{cc}4& 7\\ 3& 2\end{array}\right|i-\left|\begin{array}{cc}-9& 7\\ 6& 2\end{array}\right|j+\left|\begin{array}{cc}-9& 4\\ 6& 3\end{array}\right|k\\ =\left(8-21\right)i-\left(-18-42\right)j+\left(-27-24\right)k\\ =-13i+60j-51k\text{or}\langle -13,60,-51\rangle \end{array}$

So, the cross product $u\times v$ is $-13i+60j-51k$ .

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

  $\begin{array}{c}u\cdot \left(u\times v\right)=\left(-9i+4j+7k\right)\cdot \left(-13i+60j-51k\right)\\ =\left(-9\right)\cdot \left(-13\right)+\left(4\cdot 60\right)+\left(7\right)\times \left(-51\right)\\ =117+240-357\\ =357-357\\ =0\end{array}$

and,

  $\begin{array}{c}v\cdot \left(u\times v\right)=\left(6i+3j+2k\right)\cdot \left(-13i+60j-51k\right)\\ =\left(6\right)\cdot \left(-13\right)+\left(3\cdot 60\right)+\left(2\right)\times \left(-51\right)\\ =-78+180-102\\ =180-180\\ =0\end{array}$

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