Consider three points P=(- ( - ),- ), Q=( ( - ), ) and R=( ( - + … - Vidyadip

MCQ

Consider three points $\mathrm{P}=(-\sin (\beta-\alpha),-\cos \beta), \mathrm{Q}=(\cos (\beta-\alpha), \sin \beta)$ and $\mathrm{R}=(\cos (\beta-\alpha+\theta), \sin (\beta-\theta))$, where $0<$ $\alpha, \beta, \theta<\frac{\pi}{4}$. Then

A
$\mathrm{P}$ lies on the line segment $\mathrm{RQ}$
B
$\mathrm{Q}$ lies on the line segment $\mathrm{PR}$
C
$\mathrm{R}$ lies on the line segment $\mathrm{QP}$
✓
$\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are non-collinear

✓

Answer

Correct option: D.

$\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are non-collinear

d
$ P \equiv(-\sin (\beta-\alpha),-\cos \beta) \equiv\left(x_1, y_1\right) $

$ Q \equiv(\cos (\beta-\alpha), \sin \beta) \equiv\left(x_2, y_2\right) $

$ \text { and } R \equiv\left(x_2 \cos \theta+x_1 \sin \theta, y_2 \cos \theta+y_1 \sin \theta\right) $

$ \text { We see that } T \equiv\left(\frac{x_2 \cos \theta+x_1 \sin \theta}{\cos \theta+\sin \theta}, \frac{y_2 \cos \theta+y_1 \sin \theta}{\cos \theta+\sin \theta}\right)$

We see that $\mathrm{T} \equiv\left(\frac{\mathrm{x}_2 \cos \theta+\mathrm{x}_1 \sin \theta}{\cos \theta+\sin \theta}, \frac{\mathrm{y}_2 \cos \theta+\mathrm{y}_1 \sin \theta}{\cos \theta+\sin \theta}\right)$

and $\mathrm{P}, \mathrm{Q}, \mathrm{T}$ are collinear $\Rightarrow P, Q, R$ are non-collinear.

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