Chapter 10.2, Problem 22E

$\text{(a)}$

To determine

To find: The component form.

Answer to Problem 22E

The answer:

The component form: $-2u+5v=(-16,29)$

Explanation of Solution

Given information:

The vector is $-2u+5v$

And

  $\begin{array}{c}u=(3,-2)\\ v=(-2,5)\end{array}$

Calculation:

Know that if $v=(v_1,v_2\text{ )}$ and $k$ is a real number, then

  $kv=(k{v}_1,k{v}_2)$

Also know that,

  $\begin{array}{c}u=(u_1,u_2\text{) }\\ v=(v_1,v_2)\end{array}$

Then,

  $u+v=(u_1+v_1,u_2+v_2)$

In the exercise,

  $\begin{array}{c}u=(3,-2)\\ v=(-2,5)\end{array}$

Then,

  $\begin{array}{c}-2u+5v=(-2\cdot 3,-2\cdot (-2))+(5\cdot (-2),5\cdot 5)\\ =(-6,4)+(-10,25)\\ =(-6+(-10),4+25)\\ =(-16,29)\end{array}$

Therefore, the required component form is $-2u+5v=(-16,29)$ .

$\text{(b)}$

To determine

To find: The magnitude of the vector.

Answer to Problem 22E

The answer:

The magnitude of the vector is: $|-2u+5v|=\sqrt{1097}$

Explanation of Solution

Given information:

The vector is $-2u+5v$

And

  $\begin{array}{c}u=(3,-2)\\ v=(-2,5)\end{array}$

Calculation:

Know that the magnitude of a vector $v=(v_1,v_2)$ is equal to:

  $\left|v\right|=\sqrt{v_1^2+v_2^2}$

Hence, from part $(a)$

  $-2u+5v=(-16,29)$ then the magnitude of the vector $-2u+5v$ is equal to:

  $\begin{array}{c}|-2u+5v|=\sqrt{{(-16)}^2+{29}^2}\\ =\sqrt{256+841}\\ =\sqrt{1097}\end{array}$

Therefore, the required magnitude of the vector is: $|-2u+5v|=\sqrt{1097}$