Question

Vector A has magnitude of 5 units and makes an angle of ${30}^{\circ }$ with the x-axis. Vector B has magnitude 10 units & make an angle of ${60}^{\circ }$ with the x-axis. Find $\overrightarrow{A}+\overrightarrow{B}$

• A. $(\frac{5\sqrt{3}}{2}+\frac{5}{2})\hat{i}+(\frac{5}{2}+5)\hat{j}$
• B. $(\frac{5\sqrt{3}}{2}+5)\hat{i}+(\frac{5}{2}+\frac{5\sqrt{3}}{2})\hat{j}$
• C. $(\frac{5}{2}+5\sqrt{3})\hat{i}+(\frac{5\sqrt{3}}{2}+5)\hat{j}$
• D. $(\frac{5\sqrt{3}}{2}+5)\hat{i}(\frac{5}{2}+5\sqrt{3})\hat{j}$

Answer 1

Answer: (A), (D)

$(\frac{5\sqrt{3}}{2}+5)\hat{i}(\frac{5}{2}+5\sqrt{3})\hat{j}$

Method (1)

$\therefore$ angle between them is ${30}^{\circ }$ using parallelogram law

$|\overrightarrow{R}|=\sqrt{125+50\sqrt{3}}$

$tan\alpha =\frac{|\overrightarrow{B}|sin\theta }{|\overrightarrow{A}|+|\overrightarrow{B}|cos\theta }$Where $\alpha$ is the angle between $|\overrightarrow{R}|$ and $|\overrightarrow{A}|$

= $\frac{10sin{30}^{\circ }}{5+10cos{30}^{\circ }}$

= $\frac{10\times \frac{1}{2}}{5+10\times \frac{\sqrt{3}}{2}}$

= $\frac{5}{5+\sqrt{3}}$

$\therefore tan\alpha =\frac{5}{5+5\sqrt{3}}$

Method 2:

Resolve both vectors along the x and the y axis

$\overrightarrow{A}=|\overrightarrow{A}|cos30\hat{i}+|\overrightarrow{A}|sin30\hat{j}$

= $5\times \frac{\sqrt{3}}{2}\hat{i}+5\times \frac{1}{2}\hat{j}$

= $\frac{5\sqrt{3}}{2}\hat{i}+\frac{5}{2}\hat{j}$

$\overrightarrow{B}=|\overrightarrow{B}|cos{60}^{\circ }\hat{i}+|\overrightarrow{B}|sin{60}^{\circ }\hat{j}$

= $10\times \frac{1}{2}\hat{i}+10\times \frac{\sqrt{3}}{2}\hat{j}$

= $5\hat{i}+5\sqrt{3}\hat{j}$

$\therefore \overrightarrow{A}+\overrightarrow{B}=(\frac{5\sqrt{3}}{2}+5)\hat{i}+(\frac{5}{2}+5\sqrt{3})\hat{j}$