Question

If the vector $OP$ in $XY$ plane whose magnitude is $\sqrt{3}$ makes an angle ${60}^o$ with $Y-$ axis, the length of the component of the vector in direction of $X-$ axis is :

• A. $1$
• B. $\sqrt{3}$
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

Vector $V=x\uparrow +y\uparrow$

$\tan {30}^0=\frac{y}{x}\Rightarrow \frac{1}{\sqrt{3}}\frac{y}{x}\Rightarrow x=\sqrt{3}y$

given $|v|=\sqrt{3}$

$\Rightarrow \sqrt{x^2+y^2}=\sqrt{3}$

$\Rightarrow \sqrt{3y^2+y^2}=\sqrt{3}$ (using (1)

$\Rightarrow 4y^2=3$

$\Rightarrow y=\frac{\sqrt{3}}{2}$

$x=\sqrt{3}y=\sqrt{3}\times \frac{\sqrt{3}}{2}$

$x=\frac{3}{2}$

$\therefore$ component of vector in $x-axis$ direction is $\frac{3}{2}$.