If a and b are perpendicular vectors, then a × (a × (a × (a × b))) is equal to (A) 1/2|a|⁴b (B) a ×

Q: Why can't we just use the magnitude formula $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$?

That formula only gives you the magnitude, not the direction. In vector algebra, we need to preserve the directionality of the vector, which is why the BAC-CAB identity is essential for handling nested cross products.

Q: Is there a shortcut to avoid the step-by-step expansion?

While there are patterns in recursive cross products, the safest and most robust method for JEE Advanced is to apply the BAC-CAB rule iteratively. It prevents sign errors and ensures you don't miss the scalar factors.

Q: What happens if the vectors are not perpendicular?

If $\vec{a} \cdot \vec{b} \neq 0$, the term $(\vec{a} \cdot \vec{b})\vec{a}$ would not vanish. The expression would become significantly more complex, involving components of both $\vec{a}$ and $\vec{b}$.

Question

If $\vec{a}$ and $\vec{b}$ are perpendicular vectors, then $\vec{a} 	imes (\vec{a} 	imes (\vec{a} 	imes (\vec{a} 	imes \vec{b})))$ is equal to

Options:

(A) $\frac{1}{2}|\vec{a}|^4\vec{b}$

(B) $\vec{a} 	imes \vec{b}$

(C) $|\vec{a}|^4\vec{b}$

(D) $\vec{0}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a problem; we are witnessing the beautiful, rhythmic dance of vector algebra. When you see a nested expression like $\vec{a} 	imes (\vec{a} 	imes (\vec{a} 	imes (\vec{a} 	imes \vec{b})))$, it is natural to feel a moment of hesitation.

However, in the world of JEE Advanced, we do not fear the knot; we untangle it with the precision of a master craftsman.

Imagine standing in a three-dimensional coordinate system. You have vector $\vec{a}$ pointing along the z-axis and vector $\vec{b}$ lying flat on the x-axis. They are perpendicular, meaning the angle between them is $90^\circ$.

Because they are perpendicular, their dot product is zero:

$$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(90^\circ) = 0$$

This simple fact is the key that will unlock the entire problem. It is the silent hero of our calculation, waiting to make complex terms vanish into thin air.

### The BAC-CAB Rule

To break down this nested structure, we need our most powerful tool: the Vector Triple Product identity. You likely know it as the BAC-CAB rule:

$$\vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$$

This identity is a bridge. It transforms a cross product, which is geometrically intuitive but algebraically difficult, into a linear combination of vectors, which is much easier to handle. We will apply this rule recursively, layer by layer, from the inside out.

### The Recursive Reduction

Let us start with the innermost layer: $\vec{v}_2 = \vec{a} 	imes (\vec{a} 	imes \vec{b})$. Applying the BAC-CAB rule, we get:

$$\vec{v}_2 = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$$

Because $\vec{a} \cdot \vec{b} = 0$, the first term disappears entirely. We are left with:

$$\vec{v}_2 = -|\vec{a}|^2\vec{b}$$

Look at the elegance of this result. The double cross product has simply scaled our original vector $\vec{b}$ by the negative square of the magnitude of $\vec{a}$.

Now, we move to the next layer: $\vec{v}_3 = \vec{a} 	imes \vec{v}_2 = \vec{a} 	imes (-|\vec{a}|^2\vec{b})$. Since $-|\vec{a}|^2$ is just a scalar, we pull it out:

$$\vec{v}_3 = -|\vec{a}|^2(\vec{a} 	imes \vec{b})$$

Finally, we reach the outermost layer: $\vec{v}_4 = \vec{a} 	imes \vec{v}_3 = \vec{a} 	imes (-|\vec{a}|^2(\vec{a} 	imes \vec{b}))$. Again, pull the scalar out:

$$\vec{v}_4 = -|\vec{a}|^2(\vec{a} 	imes (\vec{a} 	imes \vec{b}))$$

We already know that $\vec{a} 	imes (\vec{a} 	imes \vec{b}) = -|\vec{a}|^2\vec{b}$. Substituting this back in, we get:

$$\vec{v}_4 = -|\vec{a}|^2(-|\vec{a}|^2\vec{b})$$

### Final Calculation

The negatives cancel out, and we are left with the final result:

$$\vec{v}_4 = |\vec{a}|^4\vec{b}$$

We started with a terrifying nested expression, and through the systematic application of the BAC-CAB rule and the geometric insight of orthogonality, we arrived at a clean, powerful result. This is the essence of physics and mathematics—taking the complex and revealing the simple truth hidden beneath.

Keep practicing this recursive thinking; it is the hallmark of a true **JEE Advanced scholar**.

Animated Solution for Mathematics - Vector Algebra: If $a$ and $b$ are perpendicular vectors, then $a \times (a \times (a \times (a \times b)))$ is equal to

Select Answer:

Visualized Solution

The Orthogonal Vectors

The Dot Product Condition

Vector Triple Product (VTP)

Visualizing the First Cross Product

Applying VTP to the Inner Layer

Simplifying the Inner Layer

The Third Nesting Level

Pulling Out the Scalar

The Final Nesting Level

Factoring the Scalar Again

Substituting the Known Value

Final Result

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The BAC-CAB Rule

The Recursive Reduction

Final Calculation

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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 and b are perpendicular vectors, then a×(a×(a×(a×b))) is equal to