Chapter 10.3, Problem 13E

To determine

To evaluate: The cross product and show that it is orthogonal to both $\text{u}$ and $\text{v}$ .

Answer to Problem 13E

The cross product is $-17\text{i}+1\text{j}+10\text{k}$ and the verification is shown below.

Explanation of Solution

Given information:

The given vectors are,

  $\begin{array}{l}\text{u}=2\text{i}+4\text{j}+3\text{k}\\ \text{v}=-\text{i}+3\text{j}-\text{2k}\end{array}$

Calculation:

Calculate the cross product.

  $\begin{array}{c}\text{u}\times \text{v}=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 2& 4& 3\\ -1& 3& -2\end{array}\right|\\ =\left(-8-9\right)\text{i}-\left(-4+3\right)\text{j}+\left(6+4\right)\text{k}\\ =-17\text{i}+1\text{j}+10\text{k}\\ \text{=}\langle -17,1,10\rangle \end{array}$

The two vectors are orthogonal if their dot product is zero.

Calculate $\left(\text{u}\times \text{v}\right)\cdot \text{u}$ .

  $\begin{array}{c}\left(\text{u}\times \text{v}\right)\cdot \text{u}=\left(-17\right)\left(2\right)+\left(1\right)\left(4\right)+\left(10\right)\left(3\right)\\ =30+30\\ =0\end{array}$

Calculate $\left(\text{u}\times \text{v}\right)\cdot \text{v}$ .

  $\begin{array}{c}\left(\text{u}\times \text{v}\right)\cdot \text{v}=\left(-17\right)\left(-1\right)+\left(1\right)\left(3\right)+\left(10\right)\left(-2\right)\\ =20-20\\ =0\end{array}$

Since $\left(\text{u}\times \text{v}\right)\cdot \text{u}$ and $\left(\text{u}\times \text{v}\right)\cdot \text{v}$ are zero the cross product is orthogonal to both $\text{u}$ and $\text{v}$ .

Therefore, the cross product is $-17\text{i}+1\text{j}+10\text{k}$ and it is orthogonal to both $\text{u}$ and $\text{v}$ .