In a triangle PQR, let a = QR, b = RP and c = PQ. If |a| = 3, |b| = 4 and a.(c – b) / c.(a – b) = |a

Q: Why do we replace $\vec{c}$ with $-\vec{a} - \vec{b}$? Can't we just solve for $\vec{c}$?

We don't know $\vec{c}$. The goal is to reduce the number of variables to just $\vec{a}$ and $\vec{b}$, which we have magnitudes for. By using the Triangle Law, we express the unknown in terms of the knowns, effectively collapsing the complexity of the problem.

Q: Wait, how did the denominator become 7? I got stuck in the dot product expansion.

It is a classic algebraic trap! When you expand $(-\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$, factor out the negative sign to get $-(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})$. This is the difference of squares: $-(|\vec{a}|^2 - |\vec{b}|^2)$. Substituting $3^2 - 4^2 = 9 - 16 = -7$, and with the negative sign outside, it becomes $7$.

Q: Is there a geometric way to solve this without Lagrange's Identity?

You could use the definition of the cross product $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta$, but that would require finding $\cos \theta$ from the dot product and then converting to $\sin \theta$. Lagrange's Identity is the direct, elegant, and standard JEE way to bridge the gap between dot and cross products.

Question

In a triangle $PQR$, let $\vec{a} = \vec{QR}, \vec{b} = \vec{RP}$ and $\vec{c} = \vec{PQ}$. If $|\vec{a}| = 3, |\vec{b}| = 4$ and $\frac{\vec{a} \cdot (\vec{c} - \vec{b})}{\vec{c} \cdot (\vec{a} - \vec{b})} = \frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|}$, then the value of $|\vec{a} 	imes \vec{b}|^2$ is ______

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into Triangle PQR

Welcome, future engineer. Today, we are not just solving a vector problem; we are uncovering the hidden symmetry within a triangle.

When you look at a problem involving vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ forming a triangle, do not see them as abstract arrows. See them as a journey. You start at $Q$, travel to $R$ (that is $\vec{a}$), then to $P$ (that is $\vec{b}$), and finally back to $Q$ (that is $\vec{c}$).

You have returned to where you started. This is the physical essence of the Triangle Law of Vector Addition:

$$\vec{a} + \vec{b} + \vec{c} = \vec{0}$$

### Phase 1: Simplifying the Landscape

We are given a daunting fraction involving dot products. But look closely at the constraint: $\vec{c} = -(\vec{a} + \vec{b})$. This is our master key.

Whenever you see $\vec{c}$ in the expression, replace it. We know the magnitudes of $\vec{a}$ and $\vec{b}$ ($3$ and $4$ respectively), but we know nothing about $\vec{c}$. By eliminating $\vec{c}$, we reduce the problem to a battle between $\vec{a}$ and $\vec{b}$.

Let us look at the numerator: $\vec{a} \cdot (\vec{c} - \vec{b})$. Substituting our master key, we get $\vec{a} \cdot ((-\vec{a} - \vec{b}) - \vec{b})$, which simplifies to $\vec{a} \cdot (-\vec{a} - 2\vec{b})$.

Distributing the dot product, we arrive at $-|\vec{a}|^2 - 2(\vec{a} \cdot \vec{b})$. Since $|\vec{a}| = 3$, this becomes $-9 - 2(\vec{a} \cdot \vec{b})$. The numerator is now a clean, manageable expression.

### Phase 2: The Algebraic Dance

Now, let us turn our attention to the denominator: $\vec{c} \cdot (\vec{a} - \vec{b})$. Again, substitute $\vec{c} = -(\vec{a} + \vec{b})$. We get $(-\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$.

If you factor out the negative sign, you see the beauty of the difference of squares:

$$-(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = -(|\vec{a}|^2 - |\vec{b}|^2)$$

With $|\vec{a}| = 3$ and $|\vec{b}| = 4$, this is $-(9 - 16) = 7$. The denominator is simply $7$.

Equating this to the right-hand side, which is $\frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|} = \frac{3}{3+4} = \frac{3}{7}$, we get the equation:

$$\frac{-9 - 2(\vec{a} \cdot \vec{b})}{7} = \frac{3}{7}$$

The sevens cancel out, leaving us with $-9 - 2(\vec{a} \cdot \vec{b}) = 3$. Solving for the dot product, we find $\vec{a} \cdot \vec{b} = -6$. This is the hidden value that unlocks the final door.

### Phase 3: The Final Climax with Lagrange

We have the dot product, but the question asks for the square of the magnitude of the cross product, $|\vec{a} 	imes \vec{b}|^2$. This is where we invoke the legendary Lagrange's Identity. It is the bridge between the dot product and the cross product, a tool every JEE aspirant must master:

$$|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$

Substitute our known values: $|\vec{a}|^2 = 9$, $|\vec{b}|^2 = 16$, and $(\vec{a} \cdot \vec{b})^2 = (-6)^2 = 36$. The calculation is swift and satisfying:

$$|\vec{a} 	imes \vec{b}|^2 = (9)(16) - 36 = 144 - 36 = 108$$

And there it is. **$108$**. You have navigated the geometry, conquered the algebra, and utilized the identity. This is the rhythm of JEE Advanced physics and math—breaking down the complex into the fundamental.

Animated Solution for Mathematics - Vector Algebra: In a triangle $P QR$ , let $a = QR, b = R P$ and $c = P Q$ . If $∣ a ∣ = 3, ∣ b ∣ = 4$ and $\frac{a \cdot ( c - b )}{c \cdot ( a - b )} = \frac{∣ a ∣}{∣ a ∣ + ∣ b ∣}$ , then the value of $∣ a \times b ∣^{2}$ is ______

Enter Numerical Value:

Visualized Solution

Triangle Vectors Setup

Triangle Law of Addition

Evaluating the RHS

Numerator Substitution

Simplifying Numerator

Denominator Substitution

Simplifying Denominator

Solving for Dot Product

Lagrange's Identity

Final Calculation

The Sigma Insight: Vector (Cross) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

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