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Collinear Points

Consider, thr...

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,

A

$P$ lies on the line segment $R Q$

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B

$Q$ lies on the line segment $P R$

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C

$R$ lies on the line segment $Q P$

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D

$P, Q$ and $R$ are non – collinear

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Solution

The correct option is D
$P, Q$ and $R$ are non – collinear

Explanation for correct answer:
Given, three points $P = [- \sin (β - α), - \cos (β)], Q = [\cos (β - α), \sin (β)]$ and $R = [\cos (β - α + θ), \sin (β - θ)],$
We know that the area of triangle is $\frac{1}{2} | \begin{matrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ 1 & 1 & 1 \end{matrix} |$
Finding area of triangle $A, B & C$ .
$= \frac{1}{2} |\begin{matrix} - \sin (β - α) & \cos (β - α) & \cos (β - α + θ) \\ - \cos (β) & \sin (β) & \sin (β - θ) \\ 1 & 1 & 1 \end{matrix}|$
$= \frac{1}{2} [- \sin (β - α) (\sin (β) - \sin (β - θ)) + (- \cos (β - α) (- \cos (β) - \sin (β - θ))) + \cos (β - α + θ) (- \cos (β) - \sin (β))]$
clearly, $∆ \neq 0$
Area of triangle will not be zero.
Therefore, points $A, B & C$ are non-collinear.
Hence, correct answer is option $(D)$ .

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Similar questions

Q. Consider three points

P = [- sin (β - α), - cos β], Q = [cos (β - α), sin β]

R = [cos (β - α + θ), sin (β - θ)],

0 < α, β, θ < \frac{π}{4} .

0 < α < \frac{π}{2}

be a fixed angle. If

p = (cos θ, sin θ)

Q = (cos (α - θ), sin (α - θ))

then Q is obtained from P by

α, β, γ, θ ϵ R

A_{θ} (α, β, γ) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (α + θ) & sin (α + θ) & 1 cos (β + θ) & sin (β + θ) & 1 cos (γ + θ) & sin (γ + θ) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

α, β, γ

are fixed. then

y = A_{x} (α, β, γ)

p (x), q (x)

r (x)

are polynomials of degree one and

a_{1}, a_{2}

a_{3}

be real numbers, then

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} p (α_{1}) & p (α_{2}) & p (α_{3}) q (α_{1}) & q (α_{2}) & q (α_{3}) r (α_{1}) & r (α_{2}) & r (α_{3}) \end{matrix} ∣ ∣ ∣ ∣ ∣ =

cos (θ - α) = a

sin (θ - β) = b

(0 < θ - α, θ - β < \frac{π}{2})

, then prove that

{cos}^{2} (α - β) + 2 a b sin (α - β)

a^{2} - b^{2}

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