Consider three points P = (-sin(β-α), -cosβ), Q = (cos(β-α), sinβ) and R = (cos(β-α+θ), sin(β-θ)), w

Q: Wait, why do we reject the possibility of the points being collinear? Is it possible that the determinant is zero?

That is a fantastic question. To be collinear, the area of the triangle (which is proportional to the determinant) must be zero. We found that the determinant is the product of two factors: $(1 - \sin\theta - \cos\theta)$ and $\cos(2\beta - \alpha)$. For the points to be collinear, at least one of these must be zero. However, given the constraints $0 < \alpha, \beta, \theta < \frac{\pi}{4}$, we can prove that neither factor can ever be zero. Therefore, the area is never zero, and the points are strictly non-collinear.

Q: Why can't we just use the slope formula to check for collinearity?

You absolutely could! You would calculate the slope of $PQ$ and the slope of $QR$ and check if they are equal. However, with these specific trigonometric coordinates, the algebra becomes incredibly messy and prone to errors. The determinant method, combined with the hidden linear relationship, allows us to bypass the messy algebra and arrive at the answer with elegance and clarity.

Q: How did you know to look for the linear relationship $\vec{R} = (\sin\theta)\vec{P} + (\cos\theta)\vec{Q}$?

It comes from pattern recognition. When you see coordinates involving $\beta - \alpha$ and $\beta - \theta$, it is a huge hint to use compound angle identities. Once you expand $\cos(\beta - \alpha + \theta)$ and $\sin(\beta - \theta)$, the terms $x_P, x_Q, y_P,$ and $y_Q$ appear naturally. Always look for these structures in coordinate geometry problems involving trigonometry; they are rarely coincidental!

Question

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

Options:

(A) (a) $P$ lies on the line segment $RQ$

(B) (b) $Q$ lies on the line segment $PR$

(C) (c) $R$ lies on the line segment $QP$

(D) (d) $P, Q, R$ are non-collinear

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Patterns

My dear student, welcome to a problem that, at first glance, looks like a chaotic mess of trigonometric functions. We have points $P, Q,$ and $R$ defined by angles $\alpha, \beta,$ and $	heta$.

If you try to calculate the slopes directly, you will likely find yourself drowning in a sea of identities. But here is the secret: JEE Advanced problems are rarely about brute force. They are about finding the hidden, elegant structure beneath the surface.

### Phase 1: Deconstructing the Coordinates

Let us look at our points:
$P = (-\sin(\beta-\alpha), -\cos\beta)$
$Q = (\cos(\beta-\alpha), \sin\beta)$
$R = (\cos(\beta-\alpha+	heta), \sin(\beta-	heta))$

Do not be intimidated by the complexity. Let us focus on $R$. The $x$-coordinate is $\cos(\beta-\alpha+	heta)$.

Using the compound angle formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$, we can expand this as:

$$x_R = \cos(\beta-\alpha)\cos	heta - \sin(\beta-\alpha)\sin	heta$$

Look closely at the terms. $\cos(\beta-\alpha)$ is exactly $x_Q$, and $-\sin(\beta-\alpha)$ is exactly $x_P$. Thus, we have discovered that $x_R = x_Q \cos	heta + x_P \sin	heta$. This is not a coincidence; it is the heartbeat of the problem.

### Phase 2: The Hidden Vector Relationship

If we apply the same logic to the $y$-coordinate of $R$, using $\sin(A-B) = \sin A \cos B - \cos A \sin B$, we get:

$$y_R = \sin\beta\cos	heta - \cos\beta\sin	heta$$

Again, $\sin\beta$ is $y_Q$, and $-\cos\beta$ is $y_P$. So, $y_R = y_Q \cos	heta + y_P \sin	heta$.

We have just uncovered a beautiful vector relationship: $\vec{R} = (\sin	heta)\vec{P} + (\cos	heta)\vec{Q}$. This tells us that $R$ is a linear combination of $P$ and $Q$. In the world of geometry, this is a massive shortcut.

### Phase 3: The Determinant Masterclass

To check for collinearity, we use the determinant $\Delta$ of the matrix formed by the coordinates. If $\Delta = 0$, the points are collinear. We set it up as:

$$\Delta = \begin{vmatrix} x_P & y_P & 1 \ x_Q & y_Q & 1 \ x_R & y_R & 1 \end{vmatrix}$$

Instead of expanding this directly, we use the linear relationship we found. We apply the row operation $R_3 ightarrow R_3 - (\sin	heta)R_1 - (\cos	heta)R_2$.

Because of our discovery in Phase 2, the first two elements of the third row vanish! We are left with:

$$\Delta = \begin{vmatrix} x_P & y_P & 1 \ x_Q & y_Q & 1 \ 0 & 0 & 1 - \sin	heta - \cos	heta \end{vmatrix}$$

Expanding along the third row, we get $\Delta = (1 - \sin	heta - \cos	heta) \det(P, Q)$.

### Phase 4: The Final Verdict

Now, we evaluate the $2 	imes 2$ minor, $\det(P, Q) = x_P y_Q - x_Q y_P$. Substituting the coordinates, we get:

$$(-\sin(\beta-\alpha))(\sin\beta) - (\cos(\beta-\alpha))(-\cos\beta)$$

$$= \cos(\beta-\alpha)\cos\beta - \sin(\beta-\alpha)\sin\beta$$

This is the classic expansion for $\cos(A+B)$, where $A = \beta-\alpha$ and $B = \beta$. So, the minor simplifies to $\cos(2\beta-\alpha)$.

Finally, we check our factors. Given $0 < 	heta < \frac{\pi}{4}$, the sum $\sin	heta + \cos	heta$ is always greater than 1, so $1 - \sin	heta - \cos	heta 
eq 0$.

Similarly, for $0 < \alpha, \beta < \frac{\pi}{4}$, the angle $2\beta-\alpha$ stays within a range where the cosine is strictly positive. Since neither factor is zero, the determinant is non-zero. The points are **non-collinear**. We have conquered the problem with logic, not just calculation!

Animated Solution for Mathematics - Straight Lines: Consider three points $P = (- sin (β - α), - cos β)$ , $Q = (cos (β - α), sin β)$ and $R = (cos (β - α + θ), sin (β - θ))$ , where $0 < α, β, θ < \frac{π}{4}$ . Then,

Select Answer:

Visualized Solution

Introduction to Points $P, Q, R$

Determinant Condition for Collinearity

Expanding $x_{R}$

Expanding $y_{R}$

The Hidden Vector Relationship

Applying Row Operations

Simplifying the Determinant

Evaluating the $2 \times 2$ Minor

Simplifying with Trigonometric Identities

Analyzing the First Factor

Analyzing the Second Factor

Final Conclusion

The Sigma Insight: Area of Triangle

The Geometry of Hidden Patterns

Phase 1

Phase 2

Phase 3

Phase 4

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