Consider three points P=(-sin (β-α),-cos β), Q=(cos (β-α), sin β) and R=(cos (β-α+θ), sin (β-θ) , where 0 < α, β, θ < (π)/4 . Then,

Correct answer: (D) P, Q, R are non- collinear. For collinear points Δ= | -sin (β-α) -cos β 1 cos (β-α) sin β 1 cos (β-α+θ) sin (β-θ) 1 | Clearly, Δ ≠ 0 for any value of α, β, θ , hence points are non-collinear.

Q27 / 66 · JEE Advanced 2008 (Paper 2) · Mathematics

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

A
$P$ lies on the line segment $R Q$
B
$Q$ lies on the line segment $P R$
C
$R$ lies on the line segment $Q P$
Correct answer
D
$P, Q, R$ are non- collinear

Solution

For collinear points $$ \Delta=\left|\begin{array}{ccc} -\sin (\beta-\alpha) & -\cos \beta & 1 \\ \cos (\beta-\alpha) & \sin \beta & 1 \\ \cos (\beta-\alpha+\theta) & \sin (\beta-\theta) & 1 \end{array}\right| $$ Clearly, $\Delta \neq 0$ for any value of $\alpha, \beta, \theta$, hence points are non-collinear.

Hence, the correct answer is option: 'D'.

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

Solution

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