Chapter 10, Problem 18CR

a.

To determine

Tofind :the component form of the vector.

Answer to Problem 18CR

The component form of the vector is $v=\langle -5,3,1\rangle$ .

Explanation of Solution

Given information:

The initial point is $\left(2,-1,2\right)$ and terminal point is $\left(-3,2,3\right)$ .

Calculation:to find component form of vector, subtract the initial point from termial point,

  $\begin{array}{l}v=\langle -3-2,2+1,3-2\rangle \\ v=\langle -5,3,1\rangle \text{ [subtract]}\end{array}$

Thus, the component form of vector is $v=\langle -5,3,1\rangle$ .

b.

To determine

To find :the magnitude of vector.

Answer to Problem 18CR

The magnitude of vector is $\sqrt{35}$ .

Explanation of Solution

Given information:

The initial point is $\left(2,-1,2\right)$ and terminal point is $\left(-3,2,3\right)$ .

Calculation: to find the magnitude of vector, square the component of vector and then add and take square root,

  $\begin{array}{l}\left|v\right|=\sqrt{{\left(-5\right)}^2+{\left(3\right)}^2+{\left(1\right)}^2}\\ \left|v\right|=\sqrt{25+9+1}\text{ [square]}\\ \left|v\right|=\sqrt{35}\text{ [add]}\end{array}$

Thus, The magnitude of vector is $\sqrt{35}$ .

c.

To determine

To find :the unit vector in direction of vector.

Answer to Problem 18CR

The unit vector is $\langle -\frac{5}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{1}{\sqrt{35}}\rangle$ .

Explanation of Solution

Given information:

The initial point is $\left(2,-1,2\right)$ and terminal point is $\left(-3,2,3\right)$ .

Calculation: to find the unit vector divide the vector by the magnitude of the vector,

  $\begin{array}{l}u=\frac{v}{\left|v\right|}\\ u=\frac{\langle -5,3,1\rangle }{\sqrt{35}}\text{ [simplify]}\\ u=\langle -\frac{5}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{1}{\sqrt{35}}\rangle \end{array}$

Thus, The unit vector is $\langle -\frac{5}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{1}{\sqrt{35}}\rangle$ .