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Question: Consider three points \(P=\left( -\sin \left( \beta -\alpha \right),-\cos \beta \right),~Q=\left( ...

Consider three points P=(−sin⁡(β−α),−cos⁡β), Q=(cos⁡(β−α),sin⁡β)P=\left( -\sin \left( \beta -\alpha \right),-\cos \beta \right),~Q=\left( \cos \left( \beta -\alpha \right),\sin \beta \right) and R=(cos⁡(β−α+θ),sin⁡(β−θ))R=\left( \cos \left( \beta -\alpha +\theta \right),\sin \left( \beta -\theta \right) \right), where 0<α,β,θ<π40<\alpha ,\beta ,\theta <\dfrac{\pi }{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

Explanation

Solution

Hint: Check collinearity of the given 3 points by using section formula.

The given points can be rewritten in simpler terms as
P=(−sin⁡(β−α),−cos⁡β)=(x1,y1)⋯⋯⋯(i)P=\left( -\sin \left( \beta -\alpha \right),-\cos \beta \right)=\left( {{x}_{1}},{{y}_{1}} \right)\cdots \cdots \cdots \left( i \right)
Q=(cos⁡(β−α),sin⁡β)=(x2,y2)⋯⋯⋯(ii)Q=\left( \cos \left( \beta -\alpha \right),\sin \beta \right)=\left( {{x}_{2}},{{y}_{2}} \right)\cdots \cdots \cdots \left( ii \right)
Let the coordinates of the third point R=(cos⁡(β−α+θ),sin⁡(β−θ))=(x3,y3)R=\left( \cos \left( \beta -\alpha +\theta \right),\sin \left( \beta - \theta \right) \right)=\left( {{x}_{3}},{{y}_{3}} \right). The x3{{x}_{3}} coordinate can be simplified as,
x3=(cos⁡(β−α+θ))=cos⁡[(β−α)+θ]{{x}_{3}}=\left( \cos \left( \beta -\alpha +\theta \right) \right)=\cos \left[ \left( \beta -\alpha \right)+\theta \right]
Applying the expansion cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b,
x3=cos⁡[(β−α)+θ]=cos⁡(β−α)cos⁡θ−sin⁡(β−α)sin⁡θ{{x}_{3}}=\cos \left[ \left( \beta -\alpha \right)+\theta \right]=\cos \left( \beta -\alpha \right)\cos \theta -\sin \left( \beta -\alpha \right)\sin \theta
Substituting the corresponding terms from equations (i)\left( i \right) and (ii)\left( ii \right),
x3=cos⁡[(β−α)+θ]=x2cos⁡θ+x1sin⁡θ{{x}_{3}}=\cos \left[ \left( \beta -\alpha \right)+\theta \right]={{x}_{2}}\cos \theta +{{x}_{1}}\sin \theta
Now, the y3{{y}_{3}} coordinate can be simplified as,
y3=sin⁡(β−θ){{y}_{3}}=\sin \left( \beta -\theta \right)
Applying the expansion sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b,
y3=sin⁡(β−θ)=sin⁡βcos⁡θ−cos⁡βsin⁡θ{{y}_{3}}=\sin \left( \beta -\theta \right)=\sin \beta \cos \theta -\cos \beta \sin \theta
Substituting the corresponding terms from equations (i)\left( i \right) and (ii)\left( ii \right),
y3=sin⁡(β−θ)=y2cos⁡θ+y1sin⁡θ{{y}_{3}}=\sin \left( \beta -\theta \right)={{y}_{2}}\cos \theta +{{y}_{1}}\sin \theta

So, therefore the third point can be written as,
R=(x2cos⁡θ+x1sin⁡θ,y2cos⁡θ+y1sin⁡θ)⋯⋯⋯(iii)R=\left( {{x}_{2}}\cos \theta +{{x}_{1}}\sin \theta ,{{y}_{2}}\cos \theta +{{y}_{1}}\sin \theta \right)\cdots \cdots \cdots \left( iii \right)
Consider the line with endpoints PQ. Also consider the point R that lies on the line diving it in the ratio as
below,

Using the section formula, the coordinates of point R can be obtained as,
R=(x1cos⁡θ+x2sin⁡θsin⁡θ+cos⁡θ,y1cos⁡θ+y2sin⁡θsin⁡θ+cos⁡θ)R=\left( \dfrac{{{x}_{1}}\cos \theta +{{x}_{2}}\sin \theta }{\sin \theta +\cos \theta },\dfrac{{{y}_{1}}\cos \theta +{{y}_{2}}\sin \theta }{\sin \theta +\cos \theta } \right)
From equation (iii)\left( iii \right), we have the coordinates of R as (x2cos⁡θ+x1sin⁡θ,y2cos⁡θ+y1sin⁡θ)\left( {{x}_{2}}\cos \theta +{{x}_{1}}\sin \theta ,{{y}_{2}}\cos \theta +{{y}_{1}}\sin \theta \right). Comparing this with the above
coordinates, it is clear that the form of the coordinates is not the same.
Therefore, the point R will not lie on the line PQ. It means that the points P, Q and R are not collinear.
Hence, we obtain the correct answer as option (d).

Note: The problem can be solved by applying the condition for collinear points. To check if the points P,
Q and R lie on the same line, consider that point Q lies on line PR. Then, the slope of line PQ and slope of
line QR must be equal for the points to be collinear.