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: Identify the coordinates and conditions.
We have the points:
- \( P = (\cos \alpha, \sin \beta) \),
- \( Q = (\sin \alpha, \cos \beta) \),
- \( R = (0, 0) \),
where \( 0<\alpha, \beta<\frac{\pi}{4} \). Since \( \alpha, \beta \) are in \( (0, \frac{\pi}{4}) \), both \( \cos \alpha, \sin \beta, \sin \alpha, \cos \beta \) 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 method.
Three points are collinear if the area of the triangle formed by them 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 given by: \[ \text{Area} = \frac{1}{2} \left| x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) \right|. \] 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 angle addition formula: \[ \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta = \cos (\alpha + \beta). \] Since \( 0<\alpha, \beta<\frac{\pi}{4} \), \( 0<\alpha + \beta<\frac{\pi}{2} \), so \( \cos (\alpha + \beta)>0 \). Thus: \[ \text{Area} = \frac{1}{2} \cos (\alpha + \beta)>0, \] because \( \cos (\alpha + \beta) \) is positive. A non-zero area indicates the points are not collinear.
Step 3: Verify against other options.
% Option (A) \( P \) lies on \( RQ \): The line segment \( RQ \) requires \( P \) to be a convex combination of \( R \) and \( Q \), i.e., \( P = tQ + (1-t)R \) for \( 0 \leq t \leq 1 \). This would imply \( \cos \alpha = t \sin \alpha \), \( \sin \beta = t \cos \beta \), which cannot hold simultaneously for all \( \alpha, \beta \) in \( (0, \frac{\pi}{4}) \).
% Option (B) \( Q \) lies on \( PR \): Similarly, \( Q = sP + (1-s)R \) would require \( \sin \alpha = s \cos \alpha \), \( \cos \beta = s \sin \beta \), which is inconsistent.
% Option (C) \( R \) lies on \( PQ \): \( R = uP + (1-u)Q \) would require \( 0 = u \cos \alpha + (1-u) \sin \alpha \), \( 0 = u \sin \beta + (1-u) \cos \beta \), leading to contradictions unless \( u = 0, 1 \), which doesn’t hold.

Step 4: Conclude the result.
Since the area is non-zero, \( P, Q, R \) are non-collinear, confirming option (D).