Question

Consider three points \( P(\cos \alpha, \sin \beta) \), \( Q(\sin \alpha, \cos \beta) \) and \( R(0, 0) \), where \( 0<\alpha, \beta<\frac{\pi}{4} \). Then:

• A. \( P \) lies on the line segment \( RQ \).
• B. \( Q \) lies on the line segment \( PR \).
• C. \( R \) lies on the line segment \( PQ \).
• D. \( P, Q, R \) are non-collinear.

Hint:

To check collinearity, use the area formula with coordinates. A zero area indicates collinearity, while a non-zero area shows the points are non-collinear.

Answer 1

The Correct Option is D

Step 1: Define points and constraints.
The points are:
- \( P = (\cos \alpha, \sin \beta) \),
- \( Q = (\sin \alpha, \cos \beta) \),
- \( R = (0, 0) \),
where \( 0<\alpha, \beta<\frac{\pi}{4} \). Since \( \alpha, \beta \) are in \( (0, \frac{\pi}{4}) \), all trigonometric functions are positive, and \( \cos \alpha, \sin \alpha<\frac{\sqrt{2}}{2} \), \( \sin \beta, \cos \beta<\frac{\sqrt{2}}{2} \).
Step 2: Check collinearity using the area formula.
Three points are collinear if the triangle's area is zero. The area of triangle \( PQR \) with vertices \( (x_1, y_1) = (\cos \alpha, \sin \beta) \), \( (x_2, y_2) = (\sin \alpha, \cos \beta) \), \( (x_3, y_3) = (0, 0) \) is: \[ \text{Area} = \frac{1}{2} \left| x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) \right|. \] Substitute the coordinates: \[ \text{Area} = \frac{1}{2} \left| \cos \alpha (\cos \beta - 0) + \sin \alpha (0 - \sin \beta) + 0 (\sin \beta - \cos \beta) \right| = \frac{1}{2} \left| \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \right|. \] Using the cosine addition formula: \[ \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta = \cos (\alpha + \beta). \] Since \( 0<\alpha, \beta<\frac{\pi}{4} \), \( 0<\alpha + \beta<\frac{\pi}{2} \), therefore \( \cos (\alpha + \beta)>0 \). Thus: \[ \text{Area} = \frac{1}{2} \cos (\alpha + \beta)>0, \] because \( \cos (\alpha + \beta) \) is positive. A non-zero area means the points are not collinear.
Step 3: Analyze other options.
% Option (A) \( P \) on \( RQ \): The line segment \( RQ \) needs \( P = tQ + (1-t)R \) for \( 0 \leq t \leq 1 \), which implies \( \cos \alpha = t \sin \alpha \) and \( \sin \beta = t \cos \beta \), which can't simultaneously hold for all \( \alpha, \beta \) in \( (0, \frac{\pi}{4}) \).
% Option (B) \( Q \) on \( PR \): Similarly, \( Q = sP + (1-s)R \) requires \( \sin \alpha = s \cos \alpha \) and \( \cos \beta = s \sin \beta \), which is inconsistent.
% Option (C) \( R \) on \( PQ \): \( R = uP + (1-u)Q \) would require \( 0 = u \cos \alpha + (1-u) \sin \alpha \) and \( 0 = u \sin \beta + (1-u) \cos \beta \), leading to contradictions unless \( u = 0, 1 \), which doesn’t hold.

Step 4: Conclusion.
Since the area is not zero, \( P, Q, R \) are non-collinear, confirming option (D).