Question

Vector $A$ having magnitude $\sqrt{2}$ units makes an angle ${45}^{\circ }$ with horizontal and vector $B$ having magnitude $2$ units makes an angle ${30}^{\circ }$ with horizontal as shown in the figure. What is $\overrightarrow{A}-\overrightarrow{B}$?

• A. $(1+\sqrt{3})\hat{i}+2\hat{j}$
• B. $(1+\sqrt{3})\hat{i}-2\hat{j}$
• C. $(1-\sqrt{3})\hat{i}+2\hat{j}$
• D. $(1-\sqrt{3})\hat{i}$

Answer 1

The correct option is D $(1-\sqrt{3})\hat{i}$

Taking components along the axes, the vector $A$ can be written as
$\overrightarrow{A}=\left(\sqrt{2}\cos {45}^{\circ }\right)\hat{i}+\left(\sqrt{2}\sin {45}^{\circ }\right)\hat{j}$
$=\hat{i}+\hat{j}$
The vector $B$ can be written as
$\overrightarrow{B}=\left(2\cos {30}^{\circ }\right)\hat{i}+\left(2\sin {30}^{\circ }\right)\hat{j}$
$=\sqrt{3}\hat{i}+\hat{j}$

Hence,
$\overrightarrow{A}$ $-$ $\overrightarrow{B}$
$=(\hat{i}+\hat{j})-(\sqrt{3}\hat{i}+\hat{j})$
$=(1-\sqrt{3})\hat{i}$