Let λ ∈ R, a = λi + 2j - 3k, b = i - λj + 2k. If ((a + b) × (a × b)) × (a - b) = 8i - 40j - 24k, the

Q: Why can't I just calculate the cross products directly using the components of $\vec{a}$ and $\vec{b}$? It seems like brute force would work.

Brute force is a trap! While it might technically work, you would be dealing with a massive algebraic mess involving $\lambda^2$ and $\lambda^3$ terms. In JEE Advanced, the examiners test your ability to use vector identities to simplify the problem *before* doing the arithmetic. Always look for the elegant path first.

Q: I got stuck when expanding $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b}))$. How do I handle the addition inside the cross product?

Think of the cross product as a distributive operation. You can treat $(\vec{a} + \vec{b})$ as a single vector, say $\vec{u}$, and expand $\vec{u} \times (\vec{a} \times \vec{b})$. This gives you $\vec{a} \times (\vec{a} \times \vec{b}) + \vec{b} \times (\vec{a} \times \vec{b})$. Now, apply the triple product identity to each term separately. It is like breaking a large rock into smaller, manageable pebbles.

Q: Wait, why does the $\lambda$ term disappear when calculating $|\vec{a}|^2 - |\vec{b}|^2$?

That is the beauty of the problem! When you calculate $|\vec{a}|^2 = \lambda^2 + 13$ and $|\vec{b}|^2 = \lambda^2 + 5$, the $\lambda^2$ terms are identical. When you subtract them, they cancel out perfectly, leaving you with a constant. This is a huge hint that you are on the right track—if the $\lambda$ didn't cancel, you might have made a calculation error.

Question

Let $\lambda \in \mathbb{R}, \vec{a} = \lambda \hat{i} + 2\hat{j} - 3\hat{k}, \vec{b} = \hat{i} - \lambda \hat{j} + 2\hat{k}$. If $((\vec{a} + \vec{b}) 	imes (\vec{a} 	imes \vec{b})) 	imes (\vec{a} - \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$, then $|\lambda (\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b})|^2$ is equal to

Options:

(A) 140

(B) 132

(C) 144

(D) 136

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The equation $((\vec{a} + \vec{b}) 	imes (\vec{a} 	imes \vec{b})) 	imes (\vec{a} - \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$ appears daunting, but it is a classic exercise in vector identity manipulation. To solve this, we must avoid brute-force component substitution and instead rely on the properties of the vector triple product.

### The Surgical Strike

We begin by focusing on the inner expression: $\vec{V}_{inner} = (\vec{a} + \vec{b}) 	imes (\vec{a} 	imes \vec{b})$. Expanding this using the distributive property, we obtain:

$$\vec{V}_{inner} = \vec{a} 	imes (\vec{a} 	imes \vec{b}) + \vec{b} 	imes (\vec{a} 	imes \vec{b})$$

Applying the Vector Triple Product identity, $\vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$, we expand the terms:

$$\vec{V}_{inner} = ((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) + ((\vec{b} \cdot \vec{b})\vec{a} - (\vec{b} \cdot \vec{a})\vec{b})$$

Grouping the terms, we simplify the expression to:

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

### The Great Collapse

Now, we incorporate the outer cross product: $\vec{V}_{inner} 	imes (\vec{a} - \vec{b})$. Distributing this product and utilizing the facts that $\vec{a} 	imes \vec{a} = 0$, $\vec{b} 	imes \vec{b} = 0$, and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$, the expression simplifies significantly:

$$\vec{V}_{inner} 	imes (\vec{a} - \vec{b}) = (|\vec{a}|^2 - |\vec{b}|^2)(\vec{a} 	imes \vec{b})$$

Given $|\vec{a}|^2 = \lambda^2 + 13$ and $|\vec{b}|^2 = \lambda^2 + 5$, the difference is exactly $8$. Thus, the equation reduces to:

$$8(\vec{a} 	imes \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k} \implies \vec{a} 	imes \vec{b} = \hat{i} - 5\hat{j} - 3\hat{k}$$

### The Final Victory

We determine $\lambda$ by calculating the cross product via the determinant method. Comparing the $\hat{i}$ component of the resulting vector to the value $1$, we find:

$$4 - 3\lambda = 1 \implies \lambda = 1$$

We are tasked with evaluating $|\lambda (\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b})|^2$. Expanding the inner cross product yields $-2(\vec{a} 	imes \vec{b})$. Squaring this with $\lambda = 1$ gives:

$$4 |\vec{a} 	imes \vec{b}|^2 = 4(1^2 + (-5)^2 + (-3)^2) = 4(1 + 25 + 9) = 4(35)$$

The final result is **140**.

Animated Solution for Mathematics - Vector Algebra: Let $λ \in R, a = λ \hat{i} + 2 \hat{j} - 3 \hat{k}, b = \hat{i} - λ \hat{j} + 2 \hat{k}$ . If $((a + b) \times (a \times b)) \times (a - b) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$ , then $∣ λ (a + b) \times (a - b) ∣^{2}$ is equal to

Select Answer:

Visualized Solution

Defining the Vectors

Analyzing the Inner Expression

Applying Vector Triple Product

Grouping Terms

The Outer Cross Product

Expanding the Outer Cross Product

Simplifying the LHS

Calculating Magnitudes

Equating LHS and RHS

Finding $λ$ via Determinant

Solving for $λ$

Simplifying the Target Expression

Final Calculation

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Surgical Strike

The Great Collapse

The Final Victory

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