Question

If $\theta$ is acute angle and the vector $\left(\sin \theta \right)\overline{i}+\left(\cos \theta \right)\overline{j}$ is perpendicular to the vector $\overline{i}-\sqrt{3}\overline{j}$ then $\theta =$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{5}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

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

$\overline{r_1}=\left(\sin \theta \right)i+\left(\cos \theta \right)iand\overline{r_2}=i-\sqrt{3}i$

$\overline{r_1}.\overline{r_2}=0$

$\Rightarrow \sin \theta -\sqrt{3}\cos \theta$

$\Rightarrow tan\theta =\sqrt{3},\therefore \theta ={60}^0=\frac{\pi }{3}$

$So,thatcorrectoptionisD.$