Chapter 6, Problem 8RE

Solution

To find: The component form and magnitude of the vector $\overrightarrow{AB}+\overrightarrow{CD}$ .

The component form of the vector $\overrightarrow{AB}+\overrightarrow{CD}$ is $〈6,-5〉$ and the magnitude is $\sqrt{61}$ units.

Given information:

The given points are $A\left(2,-1\right)$ , $B\left(3,1\right)$ , $C\left(-4,2\right)$ , and $D\left(1,-5\right)$ .

Formula used:

The component form of a vector $\overrightarrow{XY}$ with initial point $X\left(x_1,y_1\right)$ and terminal point $Y\left(x_2,y_2\right)$ is given by $〈x_2-x_1,y_2-y_1〉$ .

The sum of two vectors $\overrightarrow{XY}〈u_1,v_1〉$ and $\overrightarrow{PQ}〈u_2,v_2〉$ is given by:

  $\overrightarrow{XY}+\overrightarrow{PQ}=〈u_1+u_2,v_1+v_2〉$

The magnitude of $\text{v}$ is the length of the arrow $\overrightarrow{XY}$ , which is found using the distance formula:

  $\left|v\right|=\sqrt{x_1{}^2+y_1{}^2}$

Calculation:

Consider the given points,

  $A\left(2,-1\right)$ , $B\left(3,1\right)$ , $C\left(-4,2\right)$ , and $D\left(1,-5\right)$ .

Find the vector $\overrightarrow{AB}$ .

  $\begin{array}{c}\overrightarrow{AB}=〈3-2,1-(-1)〉\\ =〈1,1+1〉\\ =〈1,2〉\end{array}$

Find the vector $\overrightarrow{CD}$ .

  $\begin{array}{c}\overrightarrow{CD}=〈1-(-4),-5-2〉\\ =〈1+4,-7〉\\ =〈5,-7〉\end{array}$

Add the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ .

  $\begin{array}{c}\overrightarrow{AB}+\overrightarrow{CD}=〈1+5,2+(-7)〉\\ =〈6,2-7〉\\ =〈6,-5〉\end{array}$

Find the magnitude $\left|v\right|$ of $\overrightarrow{AB}+\overrightarrow{CD}$ .

  $\begin{array}{c}\left|v\right|=\sqrt{{\left(6\right)}^2+{\left(-5\right)}^2}\\ =\sqrt{36+25}\\ =\sqrt{61}\end{array}$

Hence, the component form of the vector $\overrightarrow{AB}+\overrightarrow{CD}$ is $〈6,-5〉$ and the magnitude is $\sqrt{61}$ units.