Condition for collinearity of P = $(−\sin(β −α),−cosβ)$, Q = $(\cos(β −α),\sinβ)$ and R = $(\cos(β −α + θ),\sin(β −θ))$, where $0 < α,β,θ < π/4$

Question

The following is a problem from Straight Lines, Coordinate Geometry:

Consider three points P = $(−\sin(β −α),−cosβ)$, Q = $(\cos(β −α),\sinβ)$ and R = $(\cos(β −α + θ),\sin(β −θ))$, where $0 < α,β,θ < π/4$ . Then,

(A) P lies on the line segment RQ

(B) Q lies on the line segment PR

(C) R lies on the line segment QP

(D) P,Q,R are non-collinear.

It can be seen clearly from the options that the question is related somehow to collinearity of three points P, Q and R. Suppose A = $(x_1,y_1)$, B = $(x_2,y_2)$ and $C = (x_3,y_3)$ are three distinct points in the $xy$-plane. Then the point C lies on the line AB if and only if there exists some real number $λ$ such that $$(x_3,y_3)=\left(\frac{λx_2+x_1}{λ+1},\frac{λy_2+y_1}{λ+1}\right)$$ This follows from the fact that three points are collinear if any one divides the line segment joining the other two points in some real ratio.

For this problem, I assumed that point Q divides PR in the ratio $λ:1$ where λ is any real number.

I arrived at the following expression for λ:

$$λ=\frac{\sin(\beta -\alpha)+\cos(\beta -\alpha)}{\cos(\beta-\alpha+\theta)-\cos(\beta-\alpha)}$$

Since no further information is available about the relationship between $\alpha, \beta$ or $\theta$, I am unable to proceed further.

I don't think solving by evaluating the following determinant will work here. It seems to be time consuming to simplify:

Could you please give any alternative method of solving this problem? Is there any other property governing the collinearity of three points in two dimensions?

Answer 1

As @mathlove showed, they aren't collinear in general.

If you set $\alpha,\beta$ constant, and change $\theta$, from $0$ to $2\pi$, then $R$ start form $Q$ at $\theta=0$, and then meet the $P$ at $\theta=\pi/2$ and finally returns to $P$ again at $\theta=2\pi$. In this way it produces an ellipse that is centered at origin and inscribed in the unit square.

In case of $\alpha=0, \ \beta=\pi/4$ your ellipse become a segment that is the diameter of unit square on bisects of first and third quadrant. In this case $P=(1/\sqrt2,1/\sqrt2), \ Q=(-1/\sqrt2,-1/\sqrt2)$. Now you have:

(A) $ \qquad 3\pi/2 < \theta < 2\pi $

(B) $ \qquad \pi/2 < \theta < \pi $

(C) $ \qquad 0 < \theta < \pi/2 \qquad \& \qquad \pi < \theta < 3\pi/2 $