Chapter 10, Problem 41CR

To determine

To find: the unit vector orthogonal to $u\text{ and }v$ .

Answer to Problem 41CR

The unit vector is $-\frac{\sqrt{1267}}{1267}\left(27i+23j+3k\right)$ orthogonal to $u\text{ and }v$ .

Explanation of Solution

Given information:

The vectors are $u=-2i+3j-5k,v=10i-12j+2k$ .

Formula used: the cross product of vectors,

  $u\times v=\left(u_2{v}_3-u_3{v}_2\right)i-\left(u_1{v}_3-u_3{v}_1\right)j+\left(u_1{v}_2-u_2{v}_1\right)$

Calculation: to find the unit vector orthogonal, first find the cross product of vectors,

to find the cross product, first use the formula and then substitute and simplify,

  $\begin{array}{l}u\times v=\left(u_2{v}_3-u_3{v}_2\right)i-\left(u_1{v}_3-u_3{v}_1\right)j+\left(u_1{v}_2-u_2{v}_1\right)\\ u\times v=\left(3·2-\left(-5\right)·\left(-12\right)\right)i-\left(-2·2-\left(-5\right)·10\right)j+\left(-2·\left(-12\right)-3·10\right)k\text{ [substitute]}\\ u\times v=\left(6-60\right)i-\left(-4+50\right)j+\left(24-30\right)k\text{ [multiply]}\\ u\times v=-54i-46j-6k\text{ [simplify]}\end{array}$

Now, find $|u\times v|$ ,

  $\begin{array}{l}|u\times v|=\sqrt{{\left(-54\right)}^2+{\left(-46\right)}^2+{\left(-6\right)}^2}\\ |u\times v|=\sqrt{2916+2116+36}\text{ [multiply]}\\ |u\times v|=\sqrt{5068}\text{ [add]}\end{array}$

Now, the unit vector is,

  $\begin{array}{l}\frac{u\times v}{|u\times v|}=\frac{-54i-46j-6k}{\sqrt{5068}}\text{ [substitute]}\\ \frac{u\times v}{|u\times v|}=\frac{-2}{2\sqrt{1267}}\left(27i+23j+3k\right)\text{ [simplify]}\\ \frac{u\times v}{|u\times v|}=-\frac{\sqrt{1267}}{1267}\left(27i+23j+3k\right)\text{ [simplify]}\end{array}$

Thus, the unit vector is $-\frac{\sqrt{1267}}{1267}\left(27i+23j+3k\right)$ .