Chapter 10.2, Problem 9E

a.

To determine

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

Answer to Problem 9E

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

Explanation of Solution

Given information:

The initial and terminal points of the vector $\text{v}$ are $(-7,3,4)$ and $(-3,0,0)$ respectively.

Let, the initial point of the vector $\text{v}$$=\text{P(}{p}_1,p_2,p_3)=\text{P}(-7,3,4)$

and, the terminal point of the vector $\text{v}$$=\text{Q}(q_1,q_2,q_3)=\text{Q}(-3,0,0)$ .

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$ .

  $\begin{array}{l}\therefore v_1=-3-(-7)=-3+7=4\\ \therefore v_2=0-3=-3\\ \therefore v_3=0-4=-4\end{array}$

Hence,

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

b.

To determine

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

Answer to Problem 9E

The magnitude of the vector $\text{v}$ , $\Vert \text{v}\Vert$$=\sqrt{41}$ .

Explanation of Solution

Given information:

The initial and terminal points of the vector $\text{v}$ are $(-7,3,4)$ and $(-3,0,0)$ respectively.

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

Here,

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

  $\therefore \Vert \text{v}\Vert =\sqrt{{(4)}^2+{(-3)}^2+{(-4)}^2}$$=\sqrt{16+9+16}=\sqrt{41}$

Hence,

The magnitude of the vector $\text{v}$ , $\Vert \text{v}\Vert$$=\sqrt{41}$ .

c.

To determine

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

Answer to Problem 9E

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

  u $=\langle \frac{4}{\sqrt{41}},\frac{-3}{\sqrt{41}},\frac{-4}{\sqrt{41}}\rangle$ .

Explanation of Solution

Given information:

The initial and terminal points of the vector $\text{v}$ are $(-7,3,4)$ and $(-3,0,0)$ 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,

  $\text{v}=\langle v_1,v_2,v_3\rangle =\langle 4,-3,-4\rangle$ and, $\Vert \text{v}\Vert =$$\sqrt{41}$ .

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

  $\text{u}$$=\frac{1}{\sqrt{41}}\langle 4,-3,-4\rangle =\langle \frac{4}{\sqrt{41}},\frac{-3}{\sqrt{41}},\frac{-4}{\sqrt{41}}\rangle$

Hence,

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

  u $=\langle \frac{4}{\sqrt{41}},\frac{-3}{\sqrt{41}},\frac{-4}{\sqrt{41}}\rangle$ .