If [a × b b × c c × a] = λ [a b c]² then λ is equal to (a) 0 (b) 1 (c) 2 (d) 3

Q: Wait, why do we reject the second term in the BAC-CAB expansion? Is it always zero?

Great question! In the expansion of $(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})$, the second term involves the dot product $(\vec{b} \times \vec{c}) \cdot \vec{c}$. By definition, the cross product $\vec{b} \times \vec{c}$ produces a vector that is orthogonal (perpendicular) to the plane containing $\vec{b}$ and $\vec{c}$. Since the dot product of any two perpendicular vectors is zero, this term vanishes. It is not just 'sometimes' zero; it is a fundamental property of the cross product.

Q: Why can't we just distribute the cross product like we do with multiplication?

Vector algebra is beautiful, but it has its own rules! The cross product is distributive over addition, but it is not associative, and it does not behave like simple scalar multiplication. You cannot simply 'multiply' cross products. You must use specific identities like the Vector Triple Product (BAC-CAB) or the Scalar Triple Product definitions to manipulate them correctly.

Q: Is there a geometric way to think about this? Why does it equal the square of the box product?

Absolutely! The scalar triple product $[\vec{a} \quad \vec{b} \quad \vec{c}]$ represents the volume of the parallelepiped formed by vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$. The cross products $\vec{a} \times \vec{b}$, $\vec{b} \times \vec{c}$, and $\vec{c} \times \vec{a}$ represent the area vectors of the faces of that parallelepiped. This identity essentially relates the volume of the parallelepiped formed by these face-area vectors back to the volume of the original parallelepiped. It is a deep geometric truth!

Question

If $[\vec{a} 	imes \vec{b} \ \vec{b} 	imes \vec{c} \ \vec{c} 	imes \vec{a}] = \lambda [\vec{a} \ \vec{b} \ \vec{c}]^2$ then $\lambda$ is equal to

Options:

(A) $0$

(B) $1$

(C) $2$

(D) $3$

Sigma Percentile · Accepted Answer

### The Elegance of Vector Geometry

Welcome, fellow explorers of the mathematical universe. Today, we are going to peel back the layers of a classic JEE Advanced problem.

It might look like a dense thicket of cross products and brackets, but I promise you, beneath the surface lies a structure of incredible symmetry and beauty. We are tasked with finding the value of $\lambda$ in the equation:

$$[\vec{a} 	imes \vec{b} \quad \vec{b} 	imes \vec{c} \quad \vec{c} 	imes \vec{a}] = \lambda [\vec{a} \quad \vec{b} \quad \vec{c}]^2$$

Let us embark on this journey together.

### Phase 1: Decoding the Box Product

First, let us ground ourselves. What is a scalar triple product? It is the volume of a parallelepiped defined by three vectors.

Algebraically, for any three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$, the box product is defined as:

$$[\vec{u} \quad \vec{v} \quad \vec{w}] = \vec{u} \cdot (\vec{v} 	imes \vec{w})$$

Our left-hand side (LHS) is a bit more complex: $[\vec{a} 	imes \vec{b} \quad \vec{b} 	imes \vec{c} \quad \vec{c} 	imes \vec{a}]$. Let us treat the first cross product, $\vec{a} 	imes \vec{b}$, as our vector $\vec{u}$, the second as $\vec{v}$, and the third as $\vec{w}$.

By applying the definition, the expression becomes:

$$(\vec{a} 	imes \vec{b}) \cdot ((\vec{b} 	imes \vec{c}) 	imes (\vec{c} 	imes \vec{a}))$$

Take a deep breath. It looks intimidating, but we have a powerful tool for this.

### Phase 2: The BAC-CAB Magic

Now, we focus on the inner part: $(\vec{b} 	imes \vec{c}) 	imes (\vec{c} 	imes \vec{a})$. This is a classic Vector Triple Product (VTP).

We have a vector crossed with another cross product. The identity we need is the famous 'BAC-CAB' rule:

$$\vec{X} 	imes (\vec{Y} 	imes \vec{Z}) = (\vec{X} \cdot \vec{Z})\vec{Y} - (\vec{X} \cdot \vec{Y})\vec{Z}$$

Let us map our vectors carefully. Let $\vec{X} = (\vec{b} 	imes \vec{c})$, $\vec{Y} = \vec{c}$, and $\vec{Z} = \vec{a}$. Applying the identity, we get:

$$((\vec{b} 	imes \vec{c}) \cdot \vec{a})\vec{c} - ((\vec{b} 	imes \vec{c}) \cdot \vec{c})\vec{a}$$

### Phase 3: The Vanishing Act

Here is where the physics of vectors shines. Look at the second term: $((\vec{b} 	imes \vec{c}) \cdot \vec{c})\vec{a}$.

The vector $\vec{b} 	imes \vec{c}$ is, by definition, perpendicular to the plane containing $\vec{b}$ and $\vec{c}$. Therefore, it is perpendicular to $\vec{c}$ itself.

The dot product of any two perpendicular vectors is zero. Thus, the entire second term vanishes into thin air! We are left with only the first term: $((\vec{b} 	imes \vec{c}) \cdot \vec{a})\vec{c}$.

### Phase 4: Cyclic Symmetry

We are almost there. The term $((\vec{b} 	imes \vec{c}) \cdot \vec{a})$ is simply the scalar triple product $[\vec{b} \quad \vec{c} \quad \vec{a}]$.

Due to the cyclic property of the scalar triple product, we know that:

$$[\vec{b} \quad \vec{c} \quad \vec{a}] = [\vec{a} \quad \vec{b} \quad \vec{c}]$$

So, our inner expression simplifies beautifully to $[\vec{a} \quad \vec{b} \quad \vec{c}]\vec{c}$.

### Phase 5: The Final Assembly

Now, we bring back the piece we set aside: the dot product with $\vec{a} 	imes \vec{b}$. Our expression is now:

$$(\vec{a} 	imes \vec{b}) \cdot ([\vec{a} \quad \vec{b} \quad \vec{c}]\vec{c})$$

Since $[\vec{a} \quad \vec{b} \quad \vec{c}]$ is just a scalar, we can pull it out to the front:

$$[\vec{a} \quad \vec{b} \quad \vec{c}] ((\vec{a} 	imes \vec{b}) \cdot \vec{c})$$

And what is $(\vec{a} 	imes \vec{b}) \cdot \vec{c}$? It is the definition of the scalar triple product $[\vec{a} \quad \vec{b} \quad \vec{c}]$ all over again!

We are left with $[\vec{a} \quad \vec{b} \quad \vec{c}] 	imes [\vec{a} \quad \vec{b} \quad \vec{c}] = [\vec{a} \quad \vec{b} \quad \vec{c}]^2$. Comparing this to our original equation, we see that **$\lambda = 1$**.

What a journey! We started with a complex expression and ended with a simple, elegant identity.

Animated Solution for Mathematics - Vector Algebra: If $[a \times b b \times c c \times a] = λ [a b c]^{2}$ then $λ$ is equal to

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Visualized Solution

Problem Statement

Definition of Scalar Triple Product

Expanding the LHS

Focusing on the Inner Expression

Vector Triple Product Identity

Applying VTP to the Inner Part

Simplifying the Second Term

Using Cyclic Properties of STP

Reassembling the Main Equation

Rearranging the Terms

Final Simplification of LHS

Finding the Value of $λ$

The Sigma Insight: Scalar Triple Product

The Elegance of Vector Geometry

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If [a×b b×c c×a]=λ[a b c]2 then λ is equal to