Consider the points P = (– sin(β – α) – cosβ), Q = (cos(β – α)sinβ) and R = (cos(β – α + θ)sin(β

Engineering

Mathematics

Introduction to Determinants

Question

Consider the points P = (– sin(β – α) – cosβ), Q = (cos(β – α)sinβ) 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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Δ = |\begin{array}{lll} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}|

Put β – α = ϕ and consider the determinant

$Δ = |\begin{array}{ccc} - \sin ϕ & - \cos β & 1 \\ \cos ϕ & \sin β & 1 \\ \cos (ϕ + θ) & \sin (β - θ) & 1 \end{array}|$

Using R₃ → R₃ – cosθ R₂ – sinθR₁

$Δ = |\begin{array}{ccc} - \sin ϕ & - \cos β & 1 \\ \cos ϕ & \sin β & 1 \\ 0 & 0 & 1 - \cos θ - \sin θ \end{array}|$

= (1 – cosθ – sinθ)cos(ϕ + β)

= (1 – cosθ – sinθ)cos(2β – α)

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

As $o < θ < π / 4 \Rightarrow \frac{π}{4} < θ + \frac{π}{4} < \frac{π}{2}$

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

⇒ cos(2β – α) ≠ 0

Thus Δ ≠ 0 and the points P, Q, R are non-collinear.

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