If a = (i + j + k), a.b = 1 and a x b = j - k, then b is (a) i - j + k (b) 2j - k (c) i (d) 2i

Q: Wait, why can't we just assume $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$ and solve the system of equations?

You absolutely can! If you set $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$, you will get a system of linear equations from the dot and cross product conditions. However, the Vector Triple Product is a much more elegant and faster method. In competitive exams like JEE Advanced, time is your most precious resource. Learning to recognize when to use vector identities will save you minutes of calculation and reduce the chance of algebraic errors.

Q: Is the Vector Triple Product identity always true?

Yes, it is a fundamental identity in vector algebra. It is derived from the properties of the cross product and the dot product. It is a universal tool that works for any three vectors, and it is particularly useful when you need to isolate a vector that is involved in both dot and cross product operations.

Q: What if $\vec{a} \cdot \vec{b}$ was zero?

That is a great question! If $\vec{a} \cdot \vec{b} = 0$, it means the vectors $\vec{a}$ and $\vec{b}$ are perpendicular. The identity would then simplify to $\vec{a} \times (\vec{a} \times \vec{b}) = -|\vec{a}|^2 \vec{b}$. This would make the calculation even simpler, as the first term on the right-hand side would vanish. Always look for these simplifications!

Question

If $\vec{a} = (\hat{i} + \hat{j} + \hat{k})$, $\vec{a} \cdot \vec{b} = 1$ and $\vec{a} 	imes \vec{b} = \hat{j} - \hat{k}$, then $\vec{b}$ is

Options:

(A) $\hat{i} - \hat{j} + \hat{k}$

(B) $2\hat{j} - \hat{k}$

(C) $\hat{i}$

(D) $2\hat{i}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$. We are also provided with two specific constraints regarding an unknown vector $\vec{b}$:
1. The dot product: $\vec{a} \cdot \vec{b} = 1$
2. The cross product: $\vec{a} 	imes \vec{b} = \hat{j} - \hat{k}$

While one could solve this by assuming $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$ and solving a system of linear equations, we will utilize a more elegant approach using vector identities.

### The Master Key: The Vector Triple Product

Whenever a problem provides both the dot product and the cross product of two vectors, the **Vector Triple Product** identity is the most efficient tool. The identity is defined as:

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

This identity is powerful because it isolates the unknown vector $\vec{b}$ using known quantities. We already know $\vec{a} \cdot \vec{b} = 1$, and we can easily compute $\vec{a} \cdot \vec{a} = |\vec{a}|^2$.

### Calculating the Left-Hand Side

First, we calculate the magnitude squared of $\vec{a}$:
$$\vec{a} \cdot \vec{a} = (1)^2 + (1)^2 + (1)^2 = 3$$

Next, we evaluate the left-hand side, $\vec{a} 	imes (\vec{a} 	imes \vec{b})$, using the determinant method with $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{a} 	imes \vec{b} = 0\hat{i} + 1\hat{j} - 1\hat{k}$:

$$\vec{a} 	imes (\vec{a} 	imes \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 0 & 1 & -1 \end{vmatrix}$$

Expanding the determinant along the first row:
$$\vec{a} 	imes (\vec{a} 	imes \vec{b}) = \hat{i}((1)(-1) - (1)(1)) - \hat{j}((1)(-1) - (1)(0)) + \hat{k}((1)(1) - (1)(0))$$
$$\vec{a} 	imes (\vec{a} 	imes \vec{b}) = -2\hat{i} + \hat{j} + \hat{k}$$

### The Final Synthesis

Now, we substitute all known values back into the master identity:
$$-2\hat{i} + \hat{j} + \hat{k} = 1(\hat{i} + \hat{j} + \hat{k}) - 3\vec{b}$$

Rearranging the equation to isolate $3\vec{b}$:
$$3\vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (-2\hat{i} + \hat{j} + \hat{k})$$
$$3\vec{b} = 3\hat{i} + 0\hat{j} + 0\hat{k}$$

Dividing by $3$, we arrive at the final result:
$$\mathbf{\vec{b} = \hat{i}}$$

Animated Solution for Mathematics - Vector Algebra: If $a = (\hat{i} + \hat{j} + \hat{k})$ , $a \cdot b = 1$ and $a \times b = \hat{j} - \hat{k}$ , then $b$ is

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Visualized Solution

The Given Vectors

The Vector Triple Product

Setting up the Equation

Cross Product Setup

Expanding the Determinant

Magnitude of Vector $a$

Substituting into the Identity

Isolating the Unknown

Simplifying the Expression

Final Vector $b$

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Master Key

Calculating the Left-Hand Side

The Final Synthesis

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