Chapter 8.4, Problem 9MCQ

To determine

To find: the cross product of the vectors and then verify that the resulting vector is perpendicular to the given vectors.

Answer to Problem 9MCQ

  $\left(5,3,-7\right)$

Explanation of Solution

Given information:

  $\left(1,3,2\right)\times \left(2,-1,1\right)$

Calculation:

The cross product of the vector can be evaluated as follows:

  $\begin{array}{c}\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\begin{array}{l}\overrightarrow{i}\\ a_1\\ b_1\end{array}& \begin{array}{l}\overrightarrow{j}\\ a_2\\ b_2\end{array}& \overrightarrow{\begin{array}{l}k\\ a_3\\ b_3\end{array}}\end{array}\right|\\ \left(1,3,2\right)\times \left(2,-1,1\right)=\left|\begin{array}{ccc}\begin{array}{l}\overrightarrow{i}\\ 1\\ 2\end{array}& \begin{array}{l}\overrightarrow{j}\\ 3\\ -1\end{array}& \overrightarrow{\begin{array}{l}k\\ 2\\ 1\end{array}}\end{array}\right|\\ =\left|\begin{array}{cc}\begin{array}{l}3\\ -1\end{array}& \begin{array}{l}2\\ 1\end{array}\end{array}\right|\overrightarrow{i}-\left|\begin{array}{cc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 1\end{array}\end{array}\right|\overrightarrow{j}\text{+}\left|\begin{array}{cc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}3\\ -1\end{array}\end{array}\right|\overrightarrow{k}\\ =5\overrightarrow{i}+3\overrightarrow{j}-7\overrightarrow{k}\\ =\left(5,3,-7\right)\end{array}$

Thus the cross product the vector is $\left(5,3,-7\right)$ .

Now evaluate inner product of $\left(5,3,-7\right)$ with the other given vectors as shown below.

  $\begin{array}{c}\overrightarrow{a}\cdot \overrightarrow{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3\\ \left(5,3,-7\right)\cdot \left(1,3,2\right)=\left(5\right)\left(1\right)+\left(3\right)\left(3\right)\text{+}\left(-7\right)\left(2\right)\\ =5+9-14\\ =0\\ \\ \left(5,3,-7\right)\cdot \left(2,-1,1\right)=\left(5\right)\left(2\right)+\left(3\right)\left(-1\right)\text{+}\left(-7\right)\left(1\right)\\ =10-3-7\\ =0\end{array}$

Since the inner product of both $\left(5,3,-7\right)\cdot \left(1,3,2\right)$ and $\left(5,3,-7\right)\cdot \left(2,-1,1\right)$ is zero, so the resulting vector is perpendicular to the given vectors.