Question

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 **** on the segment RQ

(B) Q **** on the segment PR

(C) R **** in the segment QP

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

Answer 1

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 - x2y1= 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.