Question

Vector $\overrightarrow{A}$ is of length 2 cm and is inclined at ${60}^{\circ }$ above the x -axis in the first quadrant. Vector $\overrightarrow{B}$ is of length 2 cm and inclined at ${60}^{\circ }$ below the x-axis in the fourth quadrant. The difference $\overrightarrow{A}-\overrightarrow{B}$ is a vector of magnitude (cm)

• A. 2 cm along + y - axis
• B. $2\sqrt{3}$ cm along + y - axis
• C. 1 cm along - x- axis
• D. 2 cm along - x - axis

Answer 1

The correct option is B $2\sqrt{3}$ cm along + y - axis

Let us assume,
$\overrightarrow{R}=\overrightarrow{A}-\overrightarrow{B}$
$=(A\cos 60\hat{i}+A\sin 60\hat{j})-(B\cos 60\hat{i}-B\sin 60\hat{j})$
$\because |\overrightarrow{A}|=|\overrightarrow{B}|=2cm$
So, $R=(A\sin 60+B\sin 60)\hat{j}=2\sqrt{3}\hat{j}$
Hence $2\sqrt{3}cm$ along + y -axis