Let O be the origin, and OX, OY, OZ be three unit vectors in the directions of the sides QR, RP, PQ,

Q: Wait, why is the angle between $\vec{OX}$ and $\vec{OY}$ equal to $\pi - R$? Shouldn't it just be $R$?

That is a fantastic question! It is a classic trap. Remember, vectors are defined by their direction. $\vec{OX}$ is parallel to $\vec{QR}$ and $\vec{OY}$ is parallel to $\vec{RP}$. If you place these vectors tail-to-tail at the origin, you are looking at the angle between the directions of the sides. If you trace the sides of the triangle, the angle $R$ is the interior angle. The angle between the vectors, when placed tail-to-tail, corresponds to the exterior angle, which is $180^\circ - R$ or $\pi - R$. Always visualize the vectors tail-to-tail!

Q: Why can't we just use the triangle inequality here? It seems simpler.

The triangle inequality $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$ is powerful, but it usually gives you an upper bound. Here, we are looking for a minimum value of a sum of cosines, which translates to a sum of dot products. The squared magnitude identity $|\vec{a} + \vec{b} + \vec{c}|^2 \ge 0$ is the standard 'JEE-level' technique for bounding dot products because it naturally generates the cross-terms we need. It is more robust for this specific algebraic structure.

Q: Why do we assume the minimum value occurs when the sum is zero? Is it always zero?

We don't assume it is zero! We use the inequality $|\vec{OX} + \vec{OY} + \vec{OZ}|^2 \ge 0$. This tells us that the expression $3 + 2E$ must be greater than or equal to zero. This gives us the range $E \ge -\frac{3}{2}$. The minimum value is the boundary of this inequality. In the context of a triangle, this minimum is achievable when the vectors sum to zero, which corresponds to the case of an equilateral triangle.

Question

Passage:
Let O be the origin, and $\vec{OX}, \vec{OY}, \vec{OZ}$ be three unit vectors in the directions of the sides $\vec{QR}, \vec{RP}, \vec{PQ}$, respectively, of a triangle PQR.

Question 1:
$|\vec{OX} 	imes \vec{OY}| =$
(A) $\sin(P+Q)$
(B) $\sin 2R$
(C) $\sin(P+R)$
(D) $\sin(Q+R)$

Question 2:
If the triangle PQR varies, then the minimum value of $\cos(P+Q)+\cos(Q+R)+\cos(R+P)$ is
(A) $\frac{5}{3}$
(B) $\frac{3}{2}$
(C) $-\frac{3}{2}$
(D) $-\frac{5}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant dance between geometry and vector algebra. Often, students look at a triangle and see only angles and sides.

In the JEE Advanced arena, we must learn to see the hidden vector structure beneath the surface. Let us dissect this problem step by step.

### Visualizing the Parallel Universe

Imagine you are standing at the origin $O$. You have three unit vectors, $\vec{OX}$, $\vec{OY}$, and $\vec{OZ}$, pointing in the directions of the sides of a triangle $PQR$.

The problem states they are unit vectors, so $|\vec{OX}| = |\vec{OY}| = |\vec{OZ}| = 1$. Whenever you see unit vectors, you should immediately think of the dot product and the cross product, as their magnitudes will simplify to unity.

We are asked to find the magnitude of the cross product $|\vec{OX} 	imes \vec{OY}|$. Recall the definition:

$$|\vec{A} 	imes \vec{B}| = |\vec{A}||\vec{B}| \sin(	heta)$$

Here, $	heta$ is the angle between the vectors. Since $\vec{OX} \parallel \vec{QR}$ and $\vec{OY} \parallel \vec{RP}$, the angle $	heta$ between them is the exterior angle at vertex $R$, which is $\pi - R$.

Thus, $|\vec{OX} 	imes \vec{OY}| = (1)(1) \sin(\pi - R) = \sin(R)$. Using the angle sum property $P+Q+R = \pi$, we know $\sin(R) = \sin(\pi - (P+Q)) = \sin(P+Q)$.

### The Algebraic Transformation

Now, let us turn our attention to the second part: minimizing the expression $E = \cos(P+Q) + \cos(Q+R) + \cos(R+P)$.

We know $P+Q = \pi - R$, therefore $\cos(P+Q) = \cos(\pi - R) = -\cos(R)$. Applying this logic to all three terms, our expression becomes:

$$E = -(\cos(R) + \cos(P) + \cos(Q))$$

To connect this to our vectors, consider the dot product $\vec{OX} \cdot \vec{OY}$. By definition, $\vec{OX} \cdot \vec{OY} = |\vec{OX}||\vec{OY}| \cos(\pi - R) = -\cos(R)$.

Suddenly, the connection is clear. Our expression $E$ is simply the sum of the pairwise dot products of our unit vectors:

$$E = \vec{OX} \cdot \vec{OY} + \vec{OY} \cdot \vec{OZ} + \vec{OZ} \cdot \vec{OX}$$

### The Master Stroke

This is where the JEE Advanced magic happens. Whenever you are faced with a sum of dot products, you must invoke the 'Squared Magnitude Identity'.

Consider the vector sum $\vec{S} = \vec{OX} + \vec{OY} + \vec{OZ}$. We know that for any vector, the magnitude squared is non-negative: $|\vec{S}|^2 \ge 0$.

Let us expand this:

$$|\vec{OX} + \vec{OY} + \vec{OZ}|^2 = |\vec{OX}|^2 + |\vec{OY}|^2 + |\vec{OZ}|^2 + 2(\vec{OX} \cdot \vec{OY} + \vec{OY} \cdot \vec{OZ} + \vec{OZ} \cdot \vec{OX}) \ge 0$$

Since these are unit vectors, $|\vec{OX}|^2 = 1$, $|\vec{OY}|^2 = 1$, and $|\vec{OZ}|^2 = 1$. Substituting these values and our expression $E$ into the inequality, we get:

$$1 + 1 + 1 + 2E \ge 0$$
$$3 + 2E \ge 0$$

Solving for $E$, we find $2E \ge -3$, which implies $E \ge -\frac{3}{2}$.

### Final Conclusion

And there it is. The **minimum value is $-\frac{3}{2}$**.

Notice how we did not need to know the specific angles $P, Q,$ or $R$. By translating the geometric constraints into vector algebra, we bypassed the complexity of trigonometry and arrived at the solution through the sheer power of vector identities.

This is the essence of JEE Advanced mathematics—finding the most elegant path through the forest of variables. Keep practicing this mindset, and you will find that no problem is truly insurmountable.

Animated Solution for Mathematics - Vector Algebra: Comprehension Passage

$∣ O X \times O Y ∣ =$

Select Answer:

If the triangle PQR varies, then the minimum value of $cos (P + Q) + cos (Q + R) + cos (R + P)$ is

Select Answer:

Visualized Solution

$△ P QR$ and Vectors

Unit Vectors at Origin

Cross Product Magnitude

Finding the Angle $θ$

Substituting Values

Angle Sum Property

Minimizing the Expression

Simplifying Cosines

Dot Product Connection

Expression as Dot Products

Vector Sum Squared Inequality

Expanding the Square

Substituting Values

Final Conclusion

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

Visualizing the Parallel Universe

The Algebraic Transformation

The Master Stroke

Final Conclusion

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Animated Solution for Mathematics - Vector Algebra: Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣OX×OY∣=

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the triangle PQR varies, then the minimum value of cos(P+Q)+cos(Q+R)+cos(R+P) is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PQR and Vectors

Unit Vectors at Origin

Cross Product Magnitude

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Substituting Values

Angle Sum Property

Minimizing the Expression

Simplifying Cosines

Dot Product Connection

Expression as Dot Products

Vector Sum Squared Inequality

Expanding the Square

Substituting Values

Final Conclusion

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

Visualizing the Parallel Universe

The Algebraic Transformation

The Master Stroke

Final Conclusion

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$∣ O X \times O Y ∣ =$

If the triangle PQR varies, then the minimum value of $cos (P + Q) + cos (Q + R) + cos (R + P)$ is

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