Let b = i + j + λk, λ ∈ R. If a is a vector such that a × b = 13i - j - 4k and a · b + 21 = 0, then

Q: Why is $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$? Can't we just solve for components?

That is a fantastic question! While you could try to solve for components, it would lead to a system of equations that is much harder to manage. The property $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$ is a fundamental geometric truth: the cross product of two vectors creates a new vector that is orthogonal to the plane containing the original two. Using this property is not just a shortcut; it is the most elegant way to unlock the unknown variable $\lambda$ immediately.

Q: Why use the Vector Triple Product (VTP) formula? Can't I just assume $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ and solve a system of equations?

You absolutely could, but think of the VTP as a 'sniper rifle' versus a 'shotgun'. If you assume components, you end up with three variables and a system of linear equations that is prone to arithmetic errors. The VTP formula $(\vec{a} \times \vec{b}) \times \vec{b} = (\vec{a} \cdot \vec{b})\vec{b} - |\vec{b}|^2\vec{a}$ allows you to isolate $\vec{a}$ directly in one clean algebraic sweep. In JEE Advanced, efficiency and reducing the surface area for errors is the name of the game!

Q: I keep getting the wrong sign in the determinant expansion. Any tips?

The determinant expansion is where most students stumble! Remember the alternating sign pattern: $(+ - +)$. When you expand along the first row, the $\hat{j}$ component MUST have a negative sign attached to its minor. Always write it out explicitly: $\hat{i}(\dots) - \hat{j}(\dots) + \hat{k}(\dots)$. If you do that, you will never lose a sign again.

Question

Let $\vec{b} = \hat{i} + \hat{j} + \lambda\hat{k}, \lambda \in \mathbb{R}$. If $\vec{a}$ is a vector such that $\vec{a} 	imes \vec{b} = 13\hat{i} - \hat{j} - 4\hat{k}$ and $\vec{a} \cdot \vec{b} + 21 = 0$, then $(\vec{b}-\vec{a}) \cdot (\hat{k}-\hat{j}) + (\vec{b}+\vec{a}) \cdot (\hat{i}-\hat{k})$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of vector algebra. This problem is a classic JEE Advanced gem that tests your ability to visualize the geometric relationships between vectors.

We are given $\vec{b} = \hat{i} + \hat{j} + \lambda\hat{k}$ and a cross product $\vec{a} 	imes \vec{b} = 13\hat{i} - \hat{j} - 4\hat{k}$. We also know $\vec{a} \cdot \vec{b} = -21$. Our mission is to find $\vec{a}$ and evaluate a final expression.

Let us begin by unmasking the mystery of $\lambda$. The cross product $\vec{a} 	imes \vec{b}$ is a vector that stands perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$. This means that the dot product of $(\vec{a} 	imes \vec{b})$ with $\vec{b}$ must be zero.

By calculating $(13\hat{i} - \hat{j} - 4\hat{k}) \cdot (\hat{i} + \hat{j} + \lambda\hat{k}) = 0$, we get:
$$13 - 1 - 4\lambda = 0$$

Solving this gives us $\lambda = 3$. Just like that, the vector $\vec{b} = \hat{i} + \hat{j} + 3\hat{k}$ is fully revealed.

### The Power of the Vector Triple Product

Now, we face the challenge: how do we extract $\vec{a}$? We have $\vec{a} 	imes \vec{b}$ and $\vec{a} \cdot \vec{b}$, which is the perfect scenario for the **Vector Triple Product (VTP)** identity.

The VTP rule states:
$$(\vec{a} 	imes \vec{b}) 	imes \vec{b} = (\vec{a} \cdot \vec{b})\vec{b} - |\vec{b}|^2\vec{a}$$

Think of this formula as a master key. It connects the cross product, the dot product, and the magnitude of the vectors. We already know $\vec{a} \cdot \vec{b} = -21$.

We calculate the magnitude squared of $\vec{b}$ as:
$$|\vec{b}|^2 = 1^2 + 1^2 + 3^2 = 11$$

The right-hand side of our equation becomes $-21\vec{b} - 11\vec{a}$. We have reduced a complex vector problem into a simple linear equation.

### The Engine Room: Determinant Expansion

Now, we must compute the left-hand side: $(\vec{a} 	imes \vec{b}) 	imes \vec{b}$. We set up the determinant:
$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 13 & -1 & -4 \ 1 & 1 & 3 \end{vmatrix}$$

Expanding this carefully, we get:
$$\hat{i}((-1)(3) - (-4)(1)) - \hat{j}((13)(3) - (-4)(1)) + \hat{k}((13)(1) - (-1)(1))$$

This simplifies to $\hat{i}(-3 + 4) - \hat{j}(39 + 4) + \hat{k}(13 + 1)$, which results in $\hat{i} - 43\hat{j} + 14\hat{k}$. This is our LHS.

Now, we equate LHS to RHS:
$$\hat{i} - 43\hat{j} + 14\hat{k} = -21(\hat{i} + \hat{j} + 3\hat{k}) - 11\vec{a}$$

Rearranging to solve for $\vec{a}$, we find:
$$11\vec{a} = -22\hat{i} + 22\hat{j} - 77\hat{k}$$

Dividing by 11, we get:
$$\vec{a} = -2\hat{i} + 2\hat{j} - 7\hat{k}$$

### The Final Assembly

We have successfully found $\vec{a}$. The final step is to evaluate $(\vec{b}-\vec{a}) \cdot (\hat{k}-\hat{j}) + (\vec{b}+\vec{a}) \cdot (\hat{i}-\hat{k})$.

We calculate:
$$\vec{b}-\vec{a} = 3\hat{i} - \hat{j} + 10\hat{k}$$
$$\vec{b}+\vec{a} = -\hat{i} + 3\hat{j} - 4\hat{k}$$

Performing the dot products:
$$(3\hat{i} - \hat{j} + 10\hat{k}) \cdot (-\hat{j} + \hat{k}) + (-\hat{i} + 3\hat{j} - 4\hat{k}) \cdot (\hat{i} - \hat{k}) = (0 + 1 + 10) + (-1 + 0 + 4) = 11 + 3 = 14$$

The final result is **14**. You have navigated through orthogonality, the vector triple product, and determinant expansion. This is the essence of JEE Advanced mathematics—not just calculation, but the strategic application of powerful tools.

Animated Solution for Mathematics - Vector Algebra: Let $b = \hat{i} + \hat{j} + λ \hat{k}, λ \in R$ . If $a$ is a vector such that $a \times b = 13 \hat{i} - \hat{j} - 4 \hat{k}$ and $a \cdot b + 21 = 0$ , then $(b - a) \cdot (\hat{k} - \hat{j}) + (b + a) \cdot (\hat{i} - \hat{k})$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Vector Setup

The Orthogonality Principle

Setting up the Dot Product

Solving for $λ$

The Vector Triple Product Strategy

Preparing the VTP Equation

Computing the LHS Cross Product

Expanding the Determinant

Equating Both Sides

Solving for Vector $a$

Calculating $(b - a)$ and $(b + a)$

Evaluating the Final Expression

Final Result

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of the Vector Triple Product

The Engine Room

The Final Assembly

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