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Consider the points P=(- ( - ), - ), Q=( ( - ), ) and R=( (

Question

Consider the points $P = (- sin (β - α), - cos β)$ , $Q = (cos (β - α), sin β)$ and $R = (cos (β - α + θ), sin (β - θ))$ , where $0 < α, β < \frac{π}{4}$ then

P

lies on the line segment

RQ

Q

lies on the line segment

PR

R

lies on the line segment

QP

P, Q, R

are non-collinear.

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Mateo Discussed

Consider the 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 < \frac{π}{4}

then

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Text solutionVerified

Δ = ∣ ∣ x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111 ∣ ∣

Put

β - α = ϕ

and consider the determinant

Δ = ∣ ∣ - sin ϕ cos ϕ cos (ϕ + θ) - cos β sin β sin (β - θ) 111 ∣ ∣

Using

R_{3} \to R_{3} - cos θ R_{2} - sin θ R_{1}

Δ = ∣ ∣ - sin ϕ cos ϕ 0 - cos β sin β 0 11 1 - cos θ - sin θ ∣ ∣

= (1 - cos θ - sin θ) cos (ϕ + β)

= (1 - cos θ - sin θ) cos (2 β - α)

= [1 - 2 sin (θ + \frac{π}{4})] cos (2 β - α)

0 < θ < π /4 \Rightarrow \frac{π}{4} < θ + \frac{π}{4} < \frac{π}{2}

\Rightarrow \frac{1}{2} < sin (θ + \frac{π}{4}) < 1

\Rightarrow cos (2 β - α) \neq = 0

Thus

Δ \neq = 0

and the points

P, Q, R

are non-collinear.

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Question Text	Consider the points $P = (- sin (β - α), - cos β)$ , $Q = (cos (β - α), sin β)$ and $R = (cos (β - α + θ), sin (β - θ))$ , where $0 < α, β < \frac{π}{4}$ then
Answer Type	Text solution:1
Upvotes	150