Question

Consider the points $P=(-\sin (\beta -\alpha ),-\cos \beta )$, $Q=\left(\cos \right(\beta -\alpha ),\sin \beta )$ and $R=\left(\cos \right(\beta -\alpha +\theta ),\sin (\beta -\theta \left)\right)$, where $0<\alpha ,\beta <\frac{\pi }{4}$ then

• A. $P$ lies on the line segment $RQ$
• B. $Q$ lies on the line segment $PR$
• C. $R$ lies on the line segment $QP$
• D. $P,Q,R$ are non-collinear.

Answer 1

Answer: (D)

$P,Q,R$ are non-collinear.

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

Put $\beta -\alpha =\varphi$ and consider the determinant
$\Delta =\left|\begin{array}{ccc}-\sin \varphi & -\cos \beta & 1\\ \cos \varphi & \sin \beta & 1\\ \cos (\varphi +\theta )& \sin (\beta -\theta )& 1\end{array}\right|$
Using $R_3\to R_3-\cos \theta R_2-\sin \theta R_1$
$\Delta =\left|\begin{array}{ccc}-\sin \varphi & -\cos \beta & 1\\ \cos \varphi & \sin \beta & 1\\ 0& 0& 1-\cos \theta -\sin \theta \end{array}\right|$
$=(1-\cos \theta -\sin \theta )\cos (\varphi +\beta )$
$=(1-\cos \theta -\sin \theta )\cos (2\beta -\alpha )$
$=[1-\sqrt{2}\sin (\theta +\frac{\pi }{4}\left)\right]\cos (2\beta -\alpha )$
As $0<\theta <\pi /4\Rightarrow \frac{\pi }{4}<\theta +\frac{\pi }{4}<\frac{\pi }{2}$
$\Rightarrow \frac{1}{\sqrt{2}}<\sin (\theta +\frac{\pi }{4})<1$
$\Rightarrow \cos (2\beta -\alpha )\ne 0$
Thus $\Delta \ne 0$ and the points $P,Q,R$ are non-collinear.