Chapter 8, Problem 38SGA

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 38SGA

  $\left(6,10,-2\right)$

Explanation of Solution

Given information:

  $\left(5,-2,5\right)\times \left(-1,0,-3\right)$

Calculation:

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

  $\begin{array}{c}\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\begin{array}{l}\overrightarrow{i}\\ a_1\\ b_1\end{array}& \begin{array}{l}\overrightarrow{j}\\ a_2\\ b_2\end{array}& \overrightarrow{\begin{array}{l}k\\ a_3\\ b_3\end{array}}\end{array}\right|\\ \left(5,-2,5\right)\times \left(-1,0,-3\right)=\left|\begin{array}{ccc}\begin{array}{l}\overrightarrow{i}\\ 5\\ -1\end{array}& \begin{array}{l}\overrightarrow{j}\\ -2\\ 0\end{array}& \overrightarrow{\begin{array}{l}k\\ 5\\ -3\end{array}}\end{array}\right|\\ =\left|\begin{array}{cc}\begin{array}{l}-2\\ 0\end{array}& \begin{array}{l}5\\ -3\end{array}\end{array}\right|\overrightarrow{i}-\left|\begin{array}{cc}\begin{array}{l}5\\ -1\end{array}& \begin{array}{l}5\\ -3\end{array}\end{array}\right|\overrightarrow{j}\text{+}\left|\begin{array}{cc}\begin{array}{l}5\\ -1\end{array}& \begin{array}{l}-2\\ 0\end{array}\end{array}\right|\overrightarrow{k}\\ =6\overrightarrow{i}+10\overrightarrow{j}-2\overrightarrow{k}\\ =\left(6,10,-2\right)\end{array}$

Thus the cross product the vector is $\left(6,10,-2\right)$ .

Now evaluate inner product of $\left(6,10,-2\right)$ with the other given vectors as shown below.

  $\begin{array}{c}\overrightarrow{a}\cdot \overrightarrow{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3\\ \left(6,10,-2\right)\cdot \left(5,-2,5\right)=\left(6\right)\left(5\right)+\left(10\right)\left(-2\right)\text{+}\left(-2\right)\left(5\right)\\ =30-20-10\\ =0\\ \\ \left(6,10,-2\right)\cdot \left(-1,0,-3\right)=\left(6\right)\left(-1\right)+\left(10\right)\left(0\right)\text{+}\left(-2\right)\left(-3\right)\\ =-6+0+6\\ =0\end{array}$

Since the inner product of both $\left(6,10,-2\right)\cdot \left(5,-2,5\right)$ and $\left(6,10,-2\right)\cdot \left(-1,0,-3\right)$ is zero, so the resulting vector is perpendicular to the given vectors.