Question

Vector $\overrightarrow{A}$ is of length $2\text{cm}$ and makes an angle ${60}^{\circ }$ with the $x$ -axis in the first quadrant. Vector $\overrightarrow{B}$ is of length $2\text{cm}$ and ${60}^{\circ }$ below the $x$-axis in the fourth quadrant. The sum $\overrightarrow{A}+\overrightarrow{B}$ is a vector of magnitude in $\text{cm}$.

• A. $2$ along $+y$-axis
• B. $2$ along $+x$-axis
• C. $1$ along $-x$-axis
• D. $2$ along $-x$-axis

Answer 1

The correct option is B $2$ along $+x$-axis

Let us assume,
$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$
$=(A\cos {60}^{\circ }\hat{i}+A\sin {60}^{\circ }\hat{j})+(B\cos {60}^{\circ }\hat{i}-B\sin {60}^{\circ }\hat{j})$
$\because |\overrightarrow{A}|=|\overrightarrow{B}|=2\text{cm}$
So, $\overrightarrow{R}=(A\cos {60}^{\circ }+B\cos {60}^{\circ })\hat{i}=2\hat{i}$
$2\text{cm}$ along $+x$ -axis