Chapter 6.3, Problem 24E

To determine

To find: the component form and magnitude of the vector $v$ .

Answer to Problem 24E

The component form the vector $v$ is $v=\langle 12,29\rangle$ .

The magnitude of the vector $v$ is $\sqrt{985}$ .

Explanation of Solution

Given information:

The given points are

  $\left(-3,11\right)$ and $\left(9,40\right)$

Formula used:

Magnitude of a vector $u=xi+yj$ is

  $\Vert u\Vert =\sqrt{{\left(x\right)}^2+{\left(y\right)}^2}$

Calculation:

The initial point of the vector is,

  $\left(-3,11\right)$

The final point of the vector is,

  $\left(9,40\right)$

Let us assume that,

  $\begin{array}{l}a\left(-3,11\right)=\left(a_1,a_2\right)\\ b\left(9,40\right)=\left(b_1,b_2\right)\end{array}$

Then the components of the vector $v=\langle v_1,v_2\rangle$ are,

  $\begin{array}{c}{v}_1=b_1-a_1\\ =9-\left(-3\right)\\ =9+3\\ =12\end{array}$

And,

  $\begin{array}{c}{v}_2=b_2-a_2\\ =40-11\\ =29\end{array}$

Therefore, the component form the vector $v$ is,

  $v=\langle 12,29\rangle$

The magnitude of the vector is,

  $\begin{array}{c}\Vert v\Vert =\sqrt{v_1{}^2+v_2{}^2}\\ =\sqrt{{\left(12\right)}^2+{\left(29\right)}^2}\\ =\sqrt{256+676}\\ =\sqrt{985}\end{array}$