91Ó°ÊÓ

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 $\stackrel{→}{A}$and $\stackrel{→}{B}$using the given components, as shown below.

SOLVE

Given that,

$\begin{array}{l}\stackrel{→}{A}=4\widehat{i}−2\widehat{j}\\ \stackrel{→}{B}=−3\widehat{i}+5\widehat{j}\end{array}$

(a) Given that the vector

$\stackrel{→}{C}=\stackrel{→}{A}+\stackrel{→}{B}$

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

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

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

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

(c) From part (a), the vector

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

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

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

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

width="106">θ=tan−131=tan−1(3)=71.60

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