If a, b, c are vectors such that [a b c] = 4 then [a x b b x c c x a] = (a) 16 (b) 64 (c) 4 (d) 8

Q: Wait, why does $(\vec{b} \times \vec{c}) \cdot \vec{c}$ equal zero? I thought dot products were only zero if vectors were perpendicular.

You are absolutely correct! And that is the key. The cross product $\vec{b} \times \vec{c}$ is, by definition, perpendicular to both $\vec{b}$ and $\vec{c}$. Therefore, its dot product with $\vec{c}$ must be zero. It is not a coincidence; it is the geometric soul of the cross product. Whenever you see a vector dotted with its own cross-product component, it is a signal that the term will vanish.

Q: Can I just assume $\vec{a}, \vec{b}, \vec{c}$ are the unit vectors $\hat{i}, \hat{j}, \hat{k}$ to solve this?

That is a brilliant 'exam-day' strategy! If the problem holds for general vectors, it must hold for specific ones. If you set $\vec{a}=\hat{i}, \vec{b}=\hat{j}, \vec{c}=\hat{k}$, the volume is 1. The cross products become $\hat{k}, \hat{i}, \hat{j}$. Their box product is 1. Since $1^2=1$, you confirm the identity. However, for JEE Advanced, knowing the derivation is what builds the intuition for harder problems where you cannot simply pick unit vectors.

Q: Is there a faster way to remember this identity?

Yes! Think of it as the 'dual' property. The box product of the cross products is simply the square of the original box product. Just remember: $[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] = [\vec{a} \quad \vec{b} \quad \vec{c}]^2$. It is a classic result that appears often in advanced mechanics and electromagnetism. Memorize it, but understand the derivation so you can reconstruct it if you ever forget.

Question

If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $[\vec{a} 	ext{ } \vec{b} 	ext{ } \vec{c}] = 4$ then $[\vec{a} 	imes \vec{b} 	ext{ } \vec{b} 	imes \vec{c} 	ext{ } \vec{c} 	imes \vec{a}] =$

Options:

(A) 16

(B) 64

(C) 4

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to evaluate the scalar triple product of three new vectors derived from the original vectors $\vec{a}, \vec{b},$ and $\vec{c}$. We are given that the volume of the parallelepiped formed by the original vectors is:

$$[\vec{a} \quad \vec{b} \quad \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c}) = 4$$

We define the new vectors as $\vec{u} = \vec{a} 	imes \vec{b}$, $\vec{v} = \vec{b} 	imes \vec{c}$, and $\vec{w} = \vec{c} 	imes \vec{a}$. Our objective is to calculate the volume of the new parallelepiped, defined by the scalar triple product $[\vec{u} \quad \vec{v} \quad \vec{w}] = \vec{u} \cdot (\vec{v} 	imes \vec{w})$.

### The Anatomy of the Cross Product

To find the volume, we must evaluate the expression:

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

Let $\vec{m} = \vec{b} 	imes \vec{c}$. We now focus on the vector triple product $\vec{m} 	imes (\vec{c} 	imes \vec{a})$.

### The Vector Triple Product: The BAC-CAB Rule

Using the **BAC-CAB identity**, which states that $\vec{P} 	imes (\vec{Q} 	imes \vec{R}) = (\vec{P} \cdot \vec{R})\vec{Q} - (\vec{P} \cdot \vec{Q})\vec{R}$, we expand the expression:

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

Substituting $\vec{m} = \vec{b} 	imes \vec{c}$ back into the equation, we obtain:

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

### The Vanishing Act

Observe the second term: $((\vec{b} 	imes \vec{c}) \cdot \vec{c})\vec{a}$. Since the cross product $\vec{b} 	imes \vec{c}$ is by definition perpendicular to $\vec{c}$, their dot product is zero.

Consequently, the entire second term vanishes. We are left with:

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

### The Final Synthesis

Now, we substitute this result back into the original scalar triple product expression:

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

Since $[\vec{a} \quad \vec{b} \quad \vec{c}]$ is a scalar, we factor it out:

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

This simplifies to the square of the original scalar triple product:

$$[\vec{u} \quad \vec{v} \quad \vec{w}] = [\vec{a} \quad \vec{b} \quad \vec{c}]^2$$

Given that the original volume is $4$, the final volume is:

$$4^2 = \mathbf{16}$$

Animated Solution for Mathematics - Vector Algebra: If $a, b, c$ are vectors such that $[a b c] = 4$ then $[a \times b b \times c c \times a] =$

Select Answer:

Visualized Solution

Visualizing the Scalar Triple Product

Defining the Cross Product Vectors

Applying the Definition of Scalar Triple Product

Setting up the Vector Triple Product

Applying the VTP Identity

Substituting Back and Finding the Zero Term

Simplifying the Remaining Term

Substituting Back into the Main Expression

Establishing the Identity

Computing the Final Value

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Anatomy of the Cross Product

The Vector Triple Product

The Vanishing Act

The Final Synthesis

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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,c are vectors such that [a b c]=4 then [a×b b×c c×a]=