Question

Vector A has magnitude of 3 units and makes an angle of ${30}^{\circ }$ with the positive x-axis.

Vector B has magnitude of 4 units and makes an angle of ${120}^{\circ }$ with the positive x-axis.

Find the magnitude of the resultant and the angle that it makes with the positive x-axis.

$\left[ta{n}^{-1}\right(\frac{4}{3})={53}^{\circ }]$

• A. $|\overrightarrow{R}|=5$ Angle = ${53}^{\circ }$
• B. $|\overrightarrow{R}|=-5$ Angle = ${37}^{\circ }$
• C. $|\overrightarrow{R}|=5$ Angle = ${83}^{\circ }$
• D. $|\overrightarrow{R}|=-5$ Angle = ${90}^{\circ }$

Answer 1

The correct option is C

$|\overrightarrow{R}|=5$ Angle = ${83}^{\circ }$

Given |$\overrightarrow{A}$| = 3

|$\overrightarrow{B}$| = 4

Magnitude of Resultant will be |$\overrightarrow{R}$| = $\sqrt{|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2+2|\overrightarrow{A}||\overrightarrow{B}|cos\theta }$

Where $\theta$ is the angle between vector A & vector $\overrightarrow{B}\theta$ = 120 - 30 = ${90}^{\circ }$

$|\overrightarrow{R}|=\sqrt{(3{)}^2+(4{)}^2+\mathrm{2.3.4.}cos{90}^{\circ }}$

$|\overrightarrow{R}|=\sqrt{9+16+\mathrm{2.3.4.}0}$ = 5

So , magnitude of resultant is 5

Now tan $\theta$ = $\frac{|\overrightarrow{B}|sin\theta }{|\overrightarrow{A}|+|\overrightarrow{B}|cos\theta }$

$=\frac{4sin{90}^{\circ }}{3+4cos{90}^{\circ }}$

$tan\varphi =\frac{4}{3}$

$\varphi =ta{n}^{-1}\left(\frac{4}{3}\right)={53}^{\circ }$

Caution : - But this $\varphi$ is the angle that the resultant makes with the $\overrightarrow{A}$ vector and the question has asked about the angle that the resultant makes with the x - axis so it will be ${53}^{\circ }+{30}^{\circ }={83}^{\circ }$