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JEE Advanced 2008 · Determinants

Consider three points P=(−sin⁡(β−α),−cos⁡β)P = (-\sin(\beta-\alpha), -\cos\beta), Q=(cos⁡(β−α),sin⁡β)Q = (\cos(\beta-\alpha), \sin\beta) and R=(cos⁡(β−α+θ),sin⁡(β−θ))R = (\cos(\beta-\alpha+\theta), \sin(\beta-\theta)), where 0<α,β,θ<π40 < \alpha, \beta, \theta < \frac{\pi}{4}. Then,

Answer: D. P, Q, R are non-collinear

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P(−sin⁡(β−α), −cos⁡β)P\bigl(-\sin(\beta-\alpha),\ -\cos\beta\bigr) Q(cos⁡(β−α), sin⁡β),R(cos⁡(β−α+θ), sin⁡(β−θ))Q\bigl(\cos(\beta-\alpha),\ \sin\beta\bigr),\quad R\bigl(\cos(\beta-\alpha+\theta),\ \sin(\beta-\theta)\bigr) 0<α, β, θ<π40 < \alpha,\ \beta,\ \theta < \dfrac{\pi}{4} To find: which point lies on the segment, or non-collinear?\text{To find: which point lies on the segment, or non-collinear?} Plan: collinear⇔area(△PQR)=0\text{Plan: collinear} \Leftrightarrow \text{area}(\triangle PQR) = 0 Δ=(xQ−xP)(yR−yP)−(yQ−yP)(xR−xP)\Delta = (x_Q-x_P)(y_R-y_P) - (y_Q-y_P)(x_R-x_P) xQ−xP=cos⁡(β−α)+sin⁡(β−α)x_Q-x_P = \cos(\beta-\alpha) + \sin(\beta-\alpha) yQ−yP=sin⁡β+cos⁡βy_Q-y_P = \sin\beta + \cos\beta xR−xP=cos⁡(β−α+θ)+sin⁡(β−α),yR−yP=sin⁡(β−θ)+cos⁡βx_R-x_P = \cos(\beta-\alpha+\theta) + \sin(\beta-\alpha),\quad y_R-y_P = \sin(\beta-\theta) + \cos\beta p=β−α,q=βp = \beta-\alpha,\quad q = \beta Δ=(cos⁡p+sin⁡p)(sin⁡(q−θ)+cos⁡q)−(sin⁡q+cos⁡q)(cos⁡(p+θ)+sin⁡p)\Delta = (\cos p+\sin p)(\sin(q-\theta)+\cos q) - (\sin q+\cos q)(\cos(p+\theta)+\sin p) =(1−sin⁡θ−cos⁡θ)(cos⁡pcos⁡q−sin⁡psin⁡q)= (1-\sin\theta-\cos\theta)(\cos p\cos q - \sin p\sin q) cos⁡pcos⁡q−sin⁡psin⁡q=cos⁡(p+q)=cos⁡(2β−α)\cos p\cos q - \sin p\sin q = \cos(p+q) = \cos(2\beta-\alpha) Δ=cos⁡(2β−α) (1−sin⁡θ−cos⁡θ)\Delta = \cos(2\beta-\alpha)\,(1-\sin\theta-\cos\theta) 2β−α∈(−π4, π2)⇒cos⁡(2β−α)>02\beta-\alpha \in \left(-\tfrac{\pi}{4},\ \tfrac{\pi}{2}\right) \Rightarrow \cos(2\beta-\alpha) > 0 θ∈(0, π4)⇒1<sin⁡θ+cos⁡θ<2⇒1−sin⁡θ−cos⁡θ<0\theta \in \left(0,\ \tfrac{\pi}{4}\right) \Rightarrow 1 < \sin\theta+\cos\theta < \sqrt{2} \Rightarrow 1-\sin\theta-\cos\theta < 0 Δ<0⇒area≠0⇒P,Q,R are non-collinear\Delta < 0 \Rightarrow \text{area} \neq 0 \Rightarrow P, Q, R \text{ are non-collinear} option (4)\boxed{\text{option (4)}}

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