Chapter 10.2, Problem 24E

$\text{(a)}$

To determine

To find: The component form.

Answer to Problem 24E

The answer:

The component form: $-\frac{5}{13}u+\frac{12}{13}v=\left(-3,\frac{70}{13}\right)$

Explanation of Solution

Given information:

The vector is $-\frac{5}{13}u+\frac{12}{13}v$

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,

  $\begin{array}{c}-\frac{5}{13}u+\frac{12}{13}v\\ =\left(-\frac{24}{13},\frac{60}{13}\right)+\left(\frac{12}{13}\cdot (-2),\frac{12}{13}\cdot 5\right)\\ =\left(-\frac{15}{13},\frac{10}{13}\right)+\left(-\frac{24}{13},\frac{60}{13}\right)\\ =\left(-\frac{15}{13}+\left(-\frac{24}{13}\right),\frac{10}{13}+\frac{60}{13}\right)\\ =\left(-\frac{39}{13},\frac{70}{13}\right)\\ =\left(-3,\frac{70}{13}\right)\end{array}$

Therefore, the required component form is $-\frac{5}{13}u+\frac{12}{13}v=\left(-3,\frac{70}{13}\right)$ .

$\text{(b)}$

To determine

To find: The magnitude of the vector.

Answer to Problem 24E

The answer:

The magnitude of the vector is: $\left|-\frac{5}{13}u+\frac{12}{13}v\right|=\frac{\sqrt{6421}}{13}$

Explanation of Solution

Given information:

The vector is $-\frac{5}{13}u+\frac{12}{13}v$

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 $(a)$

  $-\frac{5}{13}u+\frac{12}{13}v=\left(-3,\frac{70}{13}\right)$ then the magnitude of the vector $-\frac{5}{13}u+\frac{12}{13}v$ is equal to:

  $\begin{array}{c}\left|-\frac{5}{13}u+\frac{12}{13}v\right|=\sqrt{{(-3)}^2+{\left(\frac{70}{13}\right)}^2}\\ =\sqrt{9+\frac{4900}{169}}\\ =\sqrt{\frac{9\cdot 169+4900}{169}}\\ =\frac{\sqrt{1521+4900}}{\sqrt{169}}\\ =\frac{\sqrt{6421}}{13}\end{array}$

Therefore, the required magnitude of the vector is: $\left|-\frac{5}{13}u+\frac{12}{13}v\right|=\frac{\sqrt{6421}}{13}$