Chapter 10.3, Problem 40E

To determine

To find: The unit vector orthogonal to both $u$ and $v$ if $u=\left(i-2j+2k\right)$ and $v=\left(2i-j-2k\right)$ .

Answer to Problem 40E

The unit vector orthogonal to both $u$ and $v$ is $\frac{2}{3}i+\frac{2}{3}j+\frac{1}{3}k$ .

Explanation of Solution

Given information:

The given value are $u=\left(i-2j+2k\right)$ and $v=\left(2i-j-2k\right)$ .

Calculation:

Calculate the cross product $u\times v$ .

  $\begin{array}{c}u\times v=\left(\begin{array}{ccc}i& j& k\\ 1& -2& 2\\ 2& -1& -2\end{array}\right)\\ =i\left[\left(-2\right)\left(-2\right)-\left(2\right)\left(-1\right)\right]-j\left[\left(1\right)\left(-2\right)-2\left(2\right)\right]+k\left[\left(1\right)\left(-1\right)-2\left(2\right)\right]\\ =i\left(4+2\right)-j\left(-2-4\right)+k\left(-1+4\right)\\ =6i+6j+3k\end{array}$

Calculate the dot product of $u\times v$ with $u$ and $v$ separately and if they will be equal to zero. So, they are orthogonal.

Calculate the value $\left(u\times v\right)\cdot u$ .

  $\begin{array}{c}\left(u\times v\right)\cdot u=\langle 6,6,3\rangle \cdot \langle 1,-2,2\rangle \\ =\left(6\right)\left(1\right)+\left(6\right)\left(-2\right)+\left(3\right)\left(2\right)\\ =6-12+6\\ =0\end{array}$

Calculate the value of $\left(u\times v\right)\cdot v$ .

  $\begin{array}{c}\left(u\times v\right)\cdot v=\langle 6,6,3\rangle \cdot \langle 2,-1,-2\rangle \\ =\left(6\right)\left(2\right)+\left(6\right)\left(-1\right)+\left(3\right)\left(-2\right)\\ =12-6-6\\ =0\end{array}$

Calculate the unit vector of $u\times v$ .

  $\begin{array}{c}\Vert u\times v\Vert =\sqrt{6^2+6^2+3^2}\\ =\sqrt{36+36+9}\\ =\sqrt{81}\\ =9\end{array}$

Calculate the unit vector orthogonal to both and $v$ .

  $\begin{array}{c}\frac{u\times v}{\Vert u\times v\Vert }=\frac{1}{9}\left(6i+6j+3k\right)\\ =\frac{2}{3}i+\frac{2}{3}j+\frac{1}{3}k\end{array}$

Therefore, the unit vector orthogonal to both $u$ and $v$ is $\frac{2}{3}i+\frac{2}{3}j+\frac{1}{3}k$ .