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Consider the points

P = [-sin(β - α), -cosβ]

Q = [cos(β - α),sin(β]

and R = [cos(β - α + θ), sin(β - θ)

where 0 < α,β, θ < π/4. Then

(A)  P lies on the segment RQ

(B)  Q lies on the segment PR

(C)  R lies in the segment QP

(D)  P, Q and R are non-collinear

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Correct option (D) P, Q and R are non-collinear

Explanation :

It is known that three points (x1, y1), (x2, y2) and (x3, y3) are collinear if and only if

Therefore, from R3 - R1sinθ - R2cosθ, we have

This implies that x1y2 - x2y= 0 or sinθ + cosθ  = 1. Since 0 <   θ <  π/4, sinθ + cosθ  1. Therefore, x1y2 x2y1 = 0.

This implies that

-sin(β - α)sin β + cosβcos(β - α) = 0

cos(β - α + β) = 0

Hence,2β - α = π/2 which is impossible because 0 < α and β < π/4. Therefore

2β - α ≠ π/2

Thus x1y2 - x2y1  0. Hence, the points P, Q and R are non-collinear.

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