Consider three points P = ( - sin , - cos ) , Q = ( cos , sin ) and R = ( cos , sin ) , where 0 4 . Then, [2008]

Question

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 $0 < α, β, θ < \frac{π}{4} .$

Then, [2008]

Smart Achievers · Accepted Answer

(4)

$Given: Three points P (- \sin (β - α), - \cos β), Q (\cos (β - α), \sin β)$

$R = (\cos (β - α + θ), \sin (β - θ)),$

where $0 < α, β, θ < \frac{π}{4}$

$∴$ $Δ =$ $| \begin{matrix} 1 & 1 & 1 \\ - \sin (β - α) & \cos (β - α) & \cos (β - α + θ) \\ - \cos β & \sin β & \sin (β - θ) \end{matrix} |$ $[C_{3} \to C_{3} - (C_{1} \sin θ + C_{2} \cos θ)]$

$Δ =$ $| \begin{matrix} 1 & 1 & 1 - \sin θ - \cos θ \\ - \sin (β - α) & \cos (β - α) & 0 \\ - \cos β & \sin β & 0 \end{matrix} |$

$= (1 - \sin θ - \cos θ) [\cos β \cos (β - α) - \sin β \sin (β - α)]$

$= [1 - (\sin θ + \cos θ)] \cos (2 β - α)$

$∵$ $0 < α, β, θ < \frac{π}{4},$ $∴$ $\sin θ + \cos θ \neq 1$

Also, $2 β - α < \frac{π}{2} \Rightarrow \cos (2 β - α) \neq 0$

$∴$ $Δ \neq 0 \Rightarrow$ The three given points are non-collinear.

Solution

Q.
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 $0 < α, β, θ < \frac{π}{4} .$

Then, [2008]

1 P lies on the line segment RQ

2 Q lies on the line segment PR

3 R lies on the line segment QP

4 P, Q, R are non-collinear

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