Question

If the points $(-2,0),(-1,\frac{1}{\sqrt{3}})$ and $(\cos \theta ,\sin \theta )$ are collinear, then the number of values of $\theta \in [0,2\pi ]$ :

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

Answer: (B)

$1$

Given points $(-2,0),(-1,\frac{1}{\sqrt{3}})$ and $(\cos \theta ,\sin \theta )$
For collinearity,

$\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=0$

$\left|\begin{array}{ccc}-2& 0& 1\\ -1& \frac{1}{\sqrt{3}}& 1\\ \cos \theta & \sin \theta & 1\end{array}\right|=0$
$\sin \theta -\frac{1}{\sqrt{3}}\cos \theta =\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos \theta =1$
$\Rightarrow \sin (\theta -\frac{\pi }{6})=\sin \frac{\pi }{2}$
$\Rightarrow \theta -\frac{\pi }{6}=n\pi +(-1{)}^n\frac{\pi }{2}$ ($n\in Z$)
$\Rightarrow \theta =\frac{2\pi }{3}$ (when $\theta \in [0,2\pi ],n=1$)