Chapter 10.3, Problem 15E

To determine

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

Answer to Problem 15E

The cross product is $-\frac{7}{6}\text{i}-\frac{7}{8}\text{j}+0\text{k}$ and the verification is shown below.

Explanation of Solution

Given information:

The given vectors are,

  $\begin{array}{l}\text{u}=\frac{1}{2}\text{i}-\frac{2}{3}\text{j}+\text{k}\\ \text{v}=-\frac{3}{4}\text{i}+\text{j}+\frac{1}{4}\text{k}\end{array}$

Calculation:

Calculate the cross product.

  $\begin{array}{c}\text{u}\times \text{v}=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ \frac{1}{2}& \frac{-2}{3}& 1\\ \frac{-3}{4}& 1& \frac{1}{4}\end{array}\right|\\ =\left(-\frac{1}{6}-1\right)\text{i}-\left(\frac{1}{8}+\frac{3}{4}\right)\text{j}+\left(\frac{1}{2}-\frac{1}{2}\right)\text{k}\\ =-\frac{7}{6}\text{i}-\frac{7}{8}\text{j}+0\text{k}\\ \text{=}\langle \frac{-7}{6},\frac{-7}{8},0\rangle \end{array}$

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

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

  $\begin{array}{c}\left(\text{u}\times \text{v}\right)\cdot \text{u}=\left(\frac{-7}{6}\right)\left(\frac{1}{2}\right)+\left(-\frac{7}{8}\right)\left(\frac{-2}{3}\right)+\left(-1\right)\left(1\right)\\ =\frac{-7}{6}+\frac{7}{12}\\ =0\end{array}$

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

  $\begin{array}{c}\left(\text{u}\times \text{v}\right)\cdot \text{v}=\left(\frac{-7}{6}\right)\left(\frac{-3}{4}\right)+\left(-\frac{7}{8}\right)\left(1\right)+\left(0\right)\left(\frac{1}{4}\right)\\ =\frac{7}{8}-\frac{7}{8}\\ =0\end{array}$

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 $-\frac{7}{6}\text{i}-\frac{7}{8}\text{j}+0\text{k}$ and it is orthogonal to both $\text{u}$ and $\text{v}$ .