Let $\overrightarrow{A}=4\widehat{i}-2\widehat{j}$, $\overrightarrow{B}=-3\widehat{i}+5\widehat{j}$and $C=A+B$.

a. Write vector $\overrightarrow{C}$in component form.

b. Draw a coordinate system and on it show vectors$\overrightarrow{A},\overrightarrow{B}$, and$\overrightarrow{C}$.

c. What are the magnitude and direction of vector $\overrightarrow{C}$?

VISUALIZE

Each vector is expressed as the addition of two components, one component along the $x$-direction and the other component in the y-direction. So we can use the tip-to-tail rule for vector addition and draw the vectors $\overrightarrow{A}$and $\overrightarrow{B}$using the given components, as shown below.

SOLVE

Given that,

$\begin{array}{l}\overrightarrow{A}=4\widehat{i}-2\widehat{j}\\ \overrightarrow{B}=-3\widehat{i}+5\widehat{j}\end{array}$

(a) Given that the vector

$\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$

Substitute for$\overrightarrow{A}$and$\overrightarrow{B}$from equations (1) and (2), we get

$\begin{array}{l}\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}\\ =(4\widehat{i}-2\widehat{j})+(-3\widehat{i}+5\widehat{j})\\ =1\widehat{i}+3\widehat{j}\end{array}$

Therefore, the vector $\overrightarrow{C}$in component form is $1\widehat{i}+3\widehat{j}$.

(b) Using the tip-to-tail rule for the vectors $\overrightarrow{A}$and $\overrightarrow{B}$to obtain the vector $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$as shown in the below figure.

(c) From part (a), the vector

$\overrightarrow{C}=1\widehat{i}+3\widehat{j}$

Hence the magnitude of the vector$\overrightarrow{C}$is,

$\begin{array}{l}\left|\overrightarrow{C}\right|=\sqrt{1^2+3^2}\\ =\sqrt{1+9}\\ =\sqrt{10}\end{array}$

And the direction of the vector$\overrightarrow{C}$is given by,

width="106">θ=tan−131=tan−1(3)=71.60

Hence the vector $\overrightarrow{C}$has a magnitude $\sqrt{10}$and is along the direction that makes an angle ${71.6}^{°}$with the positive $x$-axis.