Chapter 11, Problem 53RE

To determine

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

Answer to Problem 53RE

Theunit vector orthogonal to both u and v is $\left(\frac{-71\sqrt{7602}}{7602}i-\frac{44\sqrt{7602}}{7602}j+\frac{25\sqrt{7602}}{7602}k\right)$

Explanation of Solution

Given:

The following vectors u and v

  $\begin{array}{l}\text{u = –3i+2j–5k}\hfill \\ \text{v = 10i–15j+2k}\hfill \end{array}$

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

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

Considerthefollowingvectorsuandv

  $\begin{array}{l}\text{u = –3i+2j–5k}\hfill \\ \text{v = 10i–15j+2k}\hfill \end{array}$

Rewriteabovevectorsuandvincomponentformas $\text{u}=\langle –3,2,–5\rangle$ and $\text{v}=\langle 10,–15,2\rangle$

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

  $\begin{array}{l}u\times v=\left|\begin{array}{ccc}i& j& k\\ -3& 2& -5\\ 10& -15& 2\end{array}\right|\\ \begin{array}{l}\text{=i}\left[\left(\text{2}\right)\left(\text{2}\right)\text{–}\left(\text{–15}\right)\left(\text{–5}\right)\right]\text{–j}\left[\left(\text{–3}\right)\left(\text{2}\right)\text{–10}\left(\text{–5}\right)\right]\text{+ k}\left[\left(\text{–15}\right)\left(\text{–3}\right)\text{–}\left(\text{10}\right)\left(\text{2}\right)\right]\hfill \\ \text{ =i}\left(\text{4–75}\right)\text{–j}\left(\text{–6+50}\right)\text{+k}\left(\text{45–30}\right)\hfill \\ \text{ =–71i–44j+25k}\hfill \end{array}\end{array}$

Hence,thecrossproductoftwovectorsuandvincomponentformis $\langle –71,–44,25\rangle$

Toshowthat $\text{u×v}$ isorthogonaltobothuandvfindthedotproductof $\text{u×v}$ withuandvseparatelyandiftheywillbeequaltozeroitmeansthattheyareorthogonal.

Therefore,

  $\begin{array}{l}\left(\text{u×v}\right)\text{•u= }\langle \text{–71,–44,25}\rangle \text{•}\langle \text{–3,2,–5}\rangle \hfill \\ \text{ = }\left(\text{–71}\right)\left(\text{–3}\right)\text{+}\left(\text{–44}\right)\left(\text{2}\right)\text{+}\left(\text{25}\right)\left(\text{–5}\right)\hfill \\ \text{ = 213–88–125}\hfill \\ \text{ = 0}\hfill \end{array}$

Now,

  $\begin{array}{l}\left(\text{u×v}\right)\text{•v}=\langle –71,–44,25\rangle •\langle 10,–15,2\rangle \hfill \\ =\left(–71\right)\left(10\right)+\left(–44\right)\left(–15\right)+\left(25\right)\left(2\right)\hfill \\ =–710+660+50\hfill \\ =0\hfill \end{array}$

Thereforethedotproductofbothuandvwith $\text{u×v}$ is0andhence $\text{u×v}$ isorthogonaltobothuandv.

Nowcomputenormof $\text{u×v}$ thatis

  $\begin{array}{cc}\parallel u\times v\parallel & =\sqrt{{(-71)}^2+{(-44)}^2+{(25)}^2}\\ & =\sqrt{5041+1936+625}\\ & =\sqrt{7602}\end{array}$

Therefore,theunitvectororthogonaltobothuandvis

  $\begin{array}{cc}\frac{u\times v}{|u\times v\parallel }& =\frac{1}{\sqrt{7602}}(-71i-44j+25k)\\ & =\frac{\sqrt{7602}}{7602}(-71i-44j+25k)\end{array}$

Hence,theunitvectororthogonaltobothuandvis $\left(\frac{-71\sqrt{7602}}{7602}i-\frac{44\sqrt{7602}}{7602}j+\frac{25\sqrt{7602}}{7602}k\right)$