Question
Consider three points $$P = \left( { - \sin \left( {\beta - \alpha } \right), - \cos \beta } \right),$$ $$Q = \left( {\cos \left( {\beta - \alpha } \right),\sin \beta } \right)\,{\text{and}}$$ $$R = \left( {\cos \left( {\beta - \alpha + \theta } \right),\sin \left( {\beta - \theta } \right)} \right),$$ where $$0 < \alpha ,\beta ,\theta < \frac{\pi }{4}.$$ 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 :
$$P, Q, R$$ are non - collinear
Solution :
The given points are $$P \left( { - \sin \left( {\beta - \alpha } \right), - \cos \beta } \right),\,$$ $$Q \left( {\cos \left( {\beta - \alpha } \right),\sin \beta } \right)\,\,{\text{and}}$$ $$R \left( {\cos \left( {\beta - \alpha + \theta } \right),\sin \left( {\beta - \theta } \right)} \right),$$
Where $$0 < \alpha ,\beta ,\theta < \frac{\pi }{4}$$
\[\therefore \,\,\Delta = \left| \begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\
- \sin \left( {\beta - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\cos \left( {\beta - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {\beta - \alpha + \theta } \right)\\
\,\,\,\,\,\, - \cos \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \left( {\beta - \theta } \right)
\end{array} \right|\]
$${\text{Operating }}{C_3} - {C_1}\sin \theta - {C_2}\cos \theta ,\,{\text{we get}}$$
\[\Delta = \left| \begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \sin \theta - \cos \theta \\
- \sin \left( {\beta - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\cos \left( {\beta - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\
\,\,\, - \cos \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0
\end{array} \right|\]
$$\eqalign{
& = \left( {1 - \sin \theta - \cos \theta } \right)\left[ {\cos \beta \cos \left( {\beta - \alpha } \right) - \sin \beta \sin \left( {\beta - \alpha } \right)} \right] \cr
& \Rightarrow \,\,\Delta = \left[ {1 - \left( {\sin \theta + \cos \theta } \right)} \right]\cos \left( {2\beta - \alpha } \right) \cr
& \because 0 < \alpha ,\beta ,\theta < \frac{\pi }{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,\sin \theta + \cos \theta \ne 1 \cr
& {\text{Also 2}}\beta - \alpha < \frac{\pi }{4} \cr
& \Rightarrow \,\,\cos \left( {2\beta - \alpha } \right) \ne 0 \cr
& \therefore \,\,\Delta \ne 0 \cr} $$
⇒ the three points are non collinear.