Chapter 10.2, Problem 12E

a.

To determine

To find the component form of the vector $\text{v}$ .

Answer to Problem 12E

The component form of the vector $\text{v}=$$\langle -3,-4,5\rangle$ .

Explanation of Solution

Given information:

The initial and terminal points of the vector $\text{v}$ are $(-2,2,1)$ and $(-5,-2,6)$ respectively.

Let, the initial point of the vector $\text{v}$$=\text{P(}{p}_1,p_2,p_3)=\text{P}(-2,2,1)$

and, the terminal point of the vector $\text{v}$$=\text{Q}(q_1,q_2,q_3)=\text{Q}(-5,-2,6)$ .

So, the component form of the vector $\text{v}=\langle v_1,v_2,v_3\rangle$$=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle$ .

  $\begin{array}{l}\therefore v_1=-5-(-2)=-5+2=-3\\ \therefore v_2=-2-2=-4\\ \therefore v_3=6-1=5\end{array}$

Hence,

The component form of the vector $\text{v}=$$\langle -3,-4,5\rangle$ .

b.

To determine

To find the magnitude of the vector $\text{v}$ .

Answer to Problem 12E

The magnitude of the vector $\text{v}$ , $\Vert \text{v}\Vert$$=\sqrt{50}$ .

Explanation of Solution

Given information:

The initial and terminal points of the vector $\text{v}$ are $(-2,2,1)$ and $(-5,-2,6)$ respectively.

The magnitude of the vector $\text{v}$ , $\Vert \text{v}\Vert =\sqrt{v_1{}^2+v_2{}^2+v_3{}^2}$

Here,

  $v_1=-3,v_2=-4\text{ and, }{v}_3=5$ .

  $\therefore \Vert \text{v}\Vert =\sqrt{{(-3)}^2+{(-4)}^2+{(5)}^2}=\sqrt{9+16+25}=\sqrt{50}$

Hence,

The magnitude of the vector $\text{v}$ , $\Vert \text{v}\Vert$$=\sqrt{50}$ .

c.

To determine

To find a unit vector in the direction of $\text{v}$ .

Answer to Problem 12E

The unit vector in the direction of $\text{v}$ :

  u $=\langle \frac{-3}{\sqrt{50}},\frac{-4}{\sqrt{50}},\frac{5}{\sqrt{50}}\rangle$ .

Explanation of Solution

Given information:

The initial and terminal points of the vector $\text{v}$ are $(-2,2,1)$ and $(-5,-2,6)$ respectively.

Unit vector in the direction of $\text{v}$ :

  $\text{u}$$=\frac{\text{v}}{\Vert \text{v}\Vert }=\left(\frac{1}{\Vert \text{v}\Vert }\right)\langle v_1,v_2,v_3\rangle$ , $\left[\text{v}\ne 0\right]$

Here,

  $\text{v}=\langle v_1,v_2,v_3\rangle =\langle -3,-4,5\rangle$ and, $\Vert \text{v}\Vert =$$\sqrt{50}$ .

  $\therefore$ The unit vector in the direction of $\text{v}$ :

  $\text{u}$$=\frac{1}{\sqrt{50}}\langle -3,-4,5\rangle =\langle \frac{-3}{\sqrt{50}},\frac{-4}{\sqrt{50}},\frac{5}{\sqrt{50}}\rangle$

Hence,

The unit vector in the direction of $\text{v}$ :

  u $=\langle \frac{-3}{\sqrt{50}},\frac{-4}{\sqrt{50}},\frac{5}{\sqrt{50}}\rangle$ .