Chapter 10.3, Problem 16E

To determine

To find: The cross product of $u$ and $v$ and show that it is orthogonal to both $u$ and $v$ .

Answer to Problem 16E

The cross product of $u$ and $v$ is $u\times v=\langle -\frac{11}{20},-\frac{95}{250},\frac{5}{20}\rangle$ .

Explanation of Solution

Given information:

The given vectors are $u=\frac{2}{5}i-\frac{1}{4}j+\frac{1}{2}k$ and $v=-\frac{3}{5}i+j+\frac{1}{5}k$

Calculation:

Write the formula of cross product of two vectors $a=a_{1}i+a_{2}j+a_{3}k$ and $b=b_{1}i+b_{2}j+b_{3}k$ .

  $a\times b=\left|\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$ .

or,

  $a\times b=\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k$ .

Substitute the values in the above equation and simplify.

  $\begin{array}{c}u\times v=\left(-\frac{1}{4}\times \frac{1}{5}-\frac{1}{2}\times 1\right)i-\left(\frac{2}{5}\times \frac{1}{5}-\frac{1}{2}\times -\frac{3}{5}\right)j+\left\{\frac{2}{5}\times 1-\left(-\frac{1}{4}\right)\times \left(-\frac{3}{5}\right)\right\}k\\ =-\frac{11}{20}i-\frac{95}{250}j+\frac{5}{20}k\\ u\times v=\langle -\frac{11}{20},-\frac{95}{250},\frac{5}{20}\rangle \end{array}$ .

For $u\times v$ to be orthogonal to $u$ and $v$ , it must satisfy the relation below.

  $\begin{array}{l}\left(u\times v\right)\cdot u=0\\ \left(u\times v\right)\cdot v=0\end{array}$

Substitute the values and check.

  $\begin{array}{c}\left(u\times v\right)\cdot u=\left(-\frac{11}{20}\times \frac{2}{5}\right)+\left(-\frac{95}{250}\times -\frac{1}{4}\right)+\left(\frac{5}{20}\times \frac{1}{2}\right)\\ =0\\ \left(u\times v\right)\cdot v=\left(-\frac{11}{20}\times \frac{3}{5}\right)+\left(-\frac{95}{250}\times 1\right)+\left(\frac{5}{20}\times \frac{1}{5}\right)\\ =0\end{array}$

Thus it is proved that $u\times v$ is perpendicular to $u$ and $v$ .

Therefore, the cross product of $u$ and $v$ is $u\times v=\langle -\frac{11}{20},-\frac{95}{250},\frac{5}{20}\rangle$ .