Chapter 9.4, Problem 50E

(a)

To determine

The scalar multiple of $\text{v}$ has magnitude $\text{1}$ .

Answer to Problem 50E

The scalar multiple of $\text{v}$ has magnitude $\text{1}$ is $\frac{1}{m}$ .

Explanation of Solution

Given information:

The magnitude of the vector is $m$ .

Calculation:

The magnitude of the vector is $m$ .

Calculate the unit vector by multiply the vector $\text{v}$ by the scalar of $\frac{1}{m}$ .

Therefore, the scalar multiple of $\text{v}$ has magnitude $\text{1}$ is $\frac{1}{m}$ .

(b)

To determine

The scalar to change the vectors into unit vector.

Answer to Problem 50E

The scalar to change the vectors into unit vector is $\frac{1}{3},\frac{1}{10\sqrt{2}},\frac{1}{\sqrt{142}}$ .

Explanation of Solution

Given information:

The magnitude of the vector is $m$ .

Calculation:

Use the following vector.

$\text{u}=\langle 1,-2,2\rangle$

Calculate the magnitude using the formula $\left|\text{u}\right|=\sqrt{x^2+y^2+z^2}$ .

$\begin{array}{l}\left|\text{u}\right|=\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(2\right)}^2}\\ \left|\text{u}\right|=3\end{array}$

Calculate the unit vector by multiply vector $\text{u}$ by $1/3$ .

$\begin{array}{c}\frac{1}{3}\text{u}=\frac{1}{3}\langle 1,-2,2\rangle \\ \frac{1}{3}\text{u}=\langle \frac{1}{3},\frac{-2}{3},\frac{2}{3}\rangle \end{array}$

Use the following vector.

$\text{u}=\langle -6,8,-10\rangle$

Calculate the magnitude using the formula $\left|\text{u}\right|=\sqrt{x^2+y^2+z^2}$ .

$\begin{array}{l}\left|\text{u}\right|=\sqrt{{\left(6\right)}^2+{\left(-8\right)}^2+{\left(10\right)}^2}\\ \left|\text{u}\right|=10\sqrt{2}\end{array}$

Calculate the unit vector by multiply vector $\text{u}$ by $1/10\sqrt{2}$ .

$\begin{array}{c}\frac{1}{10\sqrt{2}}\text{u}=\frac{1}{10\sqrt{2}}\langle -6,8,-10\rangle \\ \frac{1}{10\sqrt{2}}\text{u}=\langle \frac{-3}{10\sqrt{2}},\frac{4}{10\sqrt{2}},\frac{-1}{10\sqrt{2}}\rangle \end{array}$

Use the following vector.

$\text{u}=\langle 6,5,9\rangle$

Calculate the magnitude using the formula $\left|\text{u}\right|=\sqrt{x^2+y^2+z^2}$ .

$\begin{array}{l}\left|\text{u}\right|=\sqrt{{\left(6\right)}^2+{\left(5\right)}^2+{\left(9\right)}^2}\\ \left|\text{u}\right|=\sqrt{142}\end{array}$

Calculate the unit vector by multiply vector $\text{u}$ by $1/\sqrt{142}$ .

$\begin{array}{c}\frac{1}{\sqrt{142}}\text{u}=\frac{1}{\sqrt{142}}\langle 6,5,9\rangle \\ \frac{1}{\sqrt{142}}\text{u}=\langle \frac{6}{\sqrt{142}},\frac{5}{\sqrt{142}},\frac{9}{\sqrt{142}}\rangle \end{array}$

Therefore, the scalar to change the vectors into unit vector is $\frac{1}{3},\frac{1}{10\sqrt{2}},\frac{1}{\sqrt{142}}$ .