Chapter 6, Problem 7RE

Solution

To find: The component form and magnitude of the vector $3\overrightarrow{AB}$ .

The component form of the vector $3\overrightarrow{AB}$ is $〈1,2〉$ and the magnitude is $3\sqrt{5}$ 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 product of vectors $\overrightarrow{XY}〈u_1,v_1〉$ and the scalar $k$ is given by:

  $k\overrightarrow{XY}=〈k{u}_1,k{v}_1〉$

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

  $\left|v\right|=\sqrt{u_1{}^2+v_1{}^2}$

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 $\overrightarrow{AB}$ .

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

Multiply the scalar $3$ with $\overrightarrow{AB}$ .

  $\begin{array}{c}3\overrightarrow{AB}=〈3\cdot 1,3\cdot 2〉\\ =〈3,6〉\end{array}$

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

  $\begin{array}{c}\left|v\right|=\sqrt{{\left(1\right)}^2+{\left(2\right)}^2}\\ =\sqrt{1+4}\\ =\sqrt{5}\end{array}$

So, the magnitude of $3\overrightarrow{AB}$ is $3\sqrt{5}$ .

Hence, the component form of the vector $3\overrightarrow{AB}$ is $〈3,6〉$ and the magnitude is $3\sqrt{5}$ units.