For a triangle ABC, let p = BC, q = CA and r = BA. If |p| = 2√3, |q| = 2 and cos θ = 1/√3, where θ i

Q: Wait, why is $\vec{p} + \vec{q} = \vec{r}$? Shouldn't it be $\vec{p} + \vec{q} + \vec{r} = 0$?

That is a classic point of confusion! Remember that vector addition depends on the direction. If you walk from B to C (vector $\vec{p}$) and then from C to A (vector $\vec{q}$), your net displacement is from B to A. Since $\vec{r}$ is defined as $\vec{BA}$, the path B $\to$ C $\to$ A is equivalent to the path B $\to$ A. Thus, $\vec{p} + \vec{q} = \vec{r}$. If you were going around the triangle in a continuous loop (B $\to$ C $\to$ A $\to$ B), then the sum would be zero, but here the directions are specific!

Q: Why does $\vec{p} \times \vec{p}$ vanish? Is it just a coincidence?

It is not a coincidence at all; it is a beautiful geometric necessity! The magnitude of a cross product is $|\vec{a}| |\vec{b}| \sin \theta$. When you cross a vector with itself, the angle $\theta$ between them is $0^\circ$. Since $\sin(0^\circ) = 0$, the entire cross product must be the zero vector. It is a powerful tool in JEE problems to simplify complex expressions instantly.

Q: Can I use the Lagrange Identity here instead of calculating the cross product magnitude directly?

You absolutely could! The Lagrange Identity, which states $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$, is a fantastic tool. However, in this specific problem, we already have the angle $\theta$ and the magnitudes, so using the definition $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$ is slightly more direct. Both methods will lead you to the same elegant result.

Question

For a triangle ABC, let $\vec{p} = \vec{BC}$, $\vec{q} = \vec{CA}$ and $\vec{r} = \vec{BA}$. If $|\vec{p}| = 2\sqrt{3}$, $|\vec{q}| = 2$ and $\cos 	heta = \frac{1}{\sqrt{3}}$, where $	heta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p} 	imes (\vec{q} - 3\vec{r})|^2 + 3|\vec{r}|^2$ is equal to :

Options:

(A) 410

(B) 340

(C) 200

(D) 220

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into Symmetry

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a geometric puzzle. In the world of JEE Advanced, vectors are not just arrows on a page—they are the language of the physical universe.

When you look at a triangle like $ABC$ with vectors $\vec{p} = \vec{BC}$, $\vec{q} = \vec{CA}$, and $\vec{r} = \vec{BA}$, you are looking at a closed loop of displacement. Let us embark on this journey to evaluate the expression $|\vec{p} 	imes (\vec{q} - 3\vec{r})|^2 + 3|\vec{r}|^2$.

### Phase 1: The Triangle Law of Addition

Before we touch any algebra, we must establish our foundation. The Triangle Law of Vector Addition is our compass.

If you stand at point $B$ and walk to $C$, you have traversed $\vec{p}$. If you then walk from $C$ to $A$, you have traversed $\vec{q}$. Your net displacement is the vector from $B$ to $A$, which is $\vec{r}$.

Therefore, we can write the fundamental relationship:

$$\vec{p} + \vec{q} = \vec{r}$$

This is not just an equation; it is the geometric truth of the triangle. By substituting $\vec{r}$ with $\vec{p} + \vec{q}$, we reduce the number of variables in our expression, which is always a strategic move in competitive exams.

### Phase 2: The Scalar Foundation

We are given $|\vec{p}| = 2\sqrt{3}$, $|\vec{q}| = 2$, and $\cos 	heta = \frac{1}{\sqrt{3}}$. To solve the final expression, we will eventually need the dot product $\vec{p} \cdot \vec{q}$.

Using the definition of the dot product, $\vec{p} \cdot \vec{q} = |\vec{p}| |\vec{q}| \cos 	heta$, we substitute our values:

$$\vec{p} \cdot \vec{q} = (2\sqrt{3})(2)\left(\frac{1}{\sqrt{3}}ight) = 4$$

This integer result is a sign that we are on the right track. Furthermore, to find $|\vec{r}|^2$, we square our triangle law equation:

$$|\vec{r}|^2 = |\vec{p} + \vec{q}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2(\vec{p} \cdot \vec{q})$$

Plugging in our values, we get $12 + 4 + 2(4) = 24$. We have now unlocked the second term of our final expression.

### Phase 3: The Cross Product Simplification

Now, let us tackle the beast: $|\vec{p} 	imes (\vec{q} - 3\vec{r})|^2$. First, substitute $\vec{r} = \vec{p} + \vec{q}$ into the bracket:

$$\vec{q} - 3(\vec{p} + \vec{q}) = \vec{q} - 3\vec{p} - 3\vec{q} = -3\vec{p} - 2\vec{q}$$

Now, we take the cross product with $\vec{p}$:

$$\vec{p} 	imes (-3\vec{p} - 2\vec{q}) = -3(\vec{p} 	imes \vec{p}) - 2(\vec{p} 	imes \vec{q})$$

Here is where the elegance of vector algebra shines. Since $\vec{p} 	imes \vec{p} = \vec{0}$, the first term vanishes into thin air! We are left with $-2(\vec{p} 	imes \vec{q})$. Taking the magnitude squared, we get:

$$|-2(\vec{p} 	imes \vec{q})|^2 = 4|\vec{p} 	imes \vec{q}|^2$$

### Phase 4: The Final Assembly

We are almost at the finish line. We need $|\vec{p} 	imes \vec{q}|^2$. We know that:

$$|\vec{p} 	imes \vec{q}|^2 = |\vec{p}|^2 |\vec{q}|^2 \sin^2 	heta$$

Since $\cos 	heta = \frac{1}{\sqrt{3}}$, we use the identity $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{1}{3} = \frac{2}{3}$. Substituting this back, we get:

$$|\vec{p} 	imes \vec{q}|^2 = (12)(4)\left(\frac{2}{3}ight) = 32$$

Finally, we combine everything into our original expression:

$$4|\vec{p} 	imes \vec{q}|^2 + 3|\vec{r}|^2 = 4(32) + 3(24) = 128 + 72 = \mathbf{200}$$

Look at that result. Through careful substitution and the application of vector properties, we have tamed a complex expression into a clean, integer answer. This is the essence of JEE Advanced physics and mathematics—not brute force, but the elegant application of fundamental laws.

Animated Solution for Mathematics - Vector Algebra: For a triangle ABC, let $p = B C$ , $q = C A$ and $r = B A$ . If $∣ p ∣ = 23$ , $∣ q ∣ = 2$ and $cos θ = \frac{1}{3}$ , where $θ$ is the angle between $p$ and $q$ , then $∣ p \times (q - 3 r) ∣^{2} + 3∣ r ∣^{2}$ is equal to :

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$ and Vectors

The Triangle Law of Addition

Calculating the Dot Product $p \cdot q$

Finding the Magnitude Squared $∣ r ∣^{2}$

Simplifying the Expression $q - 3 r$

Expanding the Cross Product

Result of the Cross Product

Finding $sin^{2} θ$

Calculating $∣ p \times q ∣^{2}$

Final Calculation

The Sigma Insight: Vector (Cross) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

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

The Triangle Law of Addition

Calculating the Dot Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p​⋅q​

Finding the Magnitude Squared .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣r∣2

Simplifying the Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q​−3r