Chapter 11, Problem 54RE

To determine

Find unit vector that is orthogonal to both the given u and v vector

Answer to Problem 54RE

Theunit vector orthogonal to both u and v is $\text{j}$ .

Explanation of Solution

Given:

The following vectors u and v

  $\begin{array}{l}\text{u = 4k}\hfill \\ \text{v = i+12k}\hfill \end{array}$

Consider the vectors in space $a=\langle x_1,y_1,z_1\rangle$ and $b=\langle x_2,y_2,z_2\rangle$ then the cross product of

  $a=\langle x_1,y_1,z_1\rangle$ and $b=\langle x_2,y_2,z_2\rangle$ will be the vector

  $\begin{array}{l}a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ =i\left[y_1\left(z_2\right)-y_2\left(z_1\right)\right]-j\left[x_1\left(z_2\right)-x_2\left(z_1\right)\right]+k\left[x_1\left(y_2\right)-x_2\left(y_1\right)\right]\end{array}$

Two vectors $\text{a}=\langle x_1,y_1,z_1\rangle$ and $\text{b}=\langle x_2,y_2,z_2\rangle$ are said to be orthogonal if the dot product of

  $\text{a}=\langle x_1,y_1,z_1\rangle$ and $\text{b}=\langle x_2,y_2,z_2\rangle$ is 0 that is if $\text{a•b=0}$

The magnitude of the vector a is:

  $\left|\text{a}\right|=\sqrt{x_1{}^2+y_1{}^2+z_1{}^2}$

Therefore, unit vector orthogonal to both $\text{a}=\langle x_1,y_1,z_1\rangle$ and $\text{b}=\langle x_2,y_2,z_2\rangle$ is

  $\text{n=}\frac{\text{a×b}}{\Vert \text{a×b}\Vert }$

Consider the following vectors u and v

  $\begin{array}{l}\text{u = 4k}\hfill \\ \text{v = i+12k}\hfill \end{array}$

Rewrite above vectors u and v in component form as $\text{u}=\langle 0,0,4\rangle$ and $\text{v}=\langle 1,0,12\rangle$

First of all, calculate $\text{u×v}$ and then show that $\text{u×v}$ is orthogonal to both u and v. Therefore,

  $\begin{array}{l}u\times v=\left|\begin{array}{ccc}i& j& k\\ 0& 0& 4\\ 1& 0& 12\end{array}\right|\\ \begin{array}{l}\text{=i}\left[\left(\text{0}\right)\left(\text{12}\right)\text{–}\left(\text{0}\right)\left(\text{4}\right)\right]\text{–j}\left[\left(\text{0}\right)\left(\text{12}\right)\text{–4}\left(\text{1}\right)\right]\text{+ k}\left[\left(\text{0}\right)\left(\text{0}\right)\text{–}\left(\text{1}\right)\left(\text{0}\right)\right]\hfill \\ \text{ =i}\left(\text{0–0}\right)\text{–j}\left(\text{0–4}\right)\text{+k}\left(\text{0–0}\right)\hfill \\ \text{ =4j}\hfill \end{array}\end{array}$

Hence, the cross product of two vectors u and v in component form is $\langle 0,4,0\rangle$

To show that $\text{u×v}$ is orthogonal to both u and v find the dot product of $\text{u×v}$ with u and v separately and if they will be equal to zero it means that they are orthogonal.

Therefore,

  $\begin{array}{l}\left(\text{u×v}\right)\text{•u}=\langle 0,4,0\rangle •\langle 0,0,4\rangle \hfill \\ =\left(0\right)\left(0\right)+\left(4\right)\left(0\right)+\left(0\right)\left(4\right)\hfill \\ =0–0–0\hfill \\ =0\hfill \end{array}$

Now,

  $\begin{array}{l}\begin{array}{l}\left(\text{u×v}\right)\text{•v}=\langle 0,4,0\rangle •\langle 1,0,12\rangle \hfill \\ =\left(0\right)\left(1\right)+\left(4\right)\left(0\right)+\left(0\right)\left(12\right)\hfill \\ =0+0+0\hfill \end{array} \\ =0\end{array}$

Therefore the dot product of both u and v with $\text{u×v}$ is 0 and hence $\text{u×v}$ is orthogonal to both u and v.

Now compute norm of $\text{u×v}$ that is

  $\begin{array}{cc}\parallel u\times v\parallel & =\sqrt{0^2+{(4)}^2+0^2}\\ & =\sqrt{16}\\ & =4\end{array}$

Therefore, the unit vector orthogonal to both u and v is

  $\begin{array}{cc}\frac{u\times v}{\parallel u\times v\parallel }& =\frac{1}{4}(4j)\\ & =j\end{array}$

Hence, the unit vector orthogonal to both u and v is $\text{j}$ .