Let `veca=2hati-hatj+3hatk, vecb=hati+hatj-4hatk` and non - zero vector `vecc` are such that `(veca xx vecb)xx vecc =veca xx(vecb xx vecc)`, then vector `vecc` may be

Options

A. `4hati-2hatj+6hatk`
B. `4hati+2hatj+6hatk`
C. `hati+hatj-hatk`
D. `hati-4hatj+hatk`

Answer

To solve the problem, we need to analyze the given vectors and the relationship between them. Let's break it down step by step.

### Step 1: Define the Vectors
We are given:
- \(\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}\)
- \(\vec{b} = \hat{i} + \hat{j} - 4\hat{k}\)

### Step 2: Calculate the Cross Product \(\vec{a} \times \vec{b}\)
To find \(\vec{a} \times \vec{b}\), we can use the determinant of a matrix formed by the unit vectors and the components of \(\vec{a}\) and \(\vec{b}\).

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 1 & 1 & -4 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -1 & 3 \\ 1 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \\ 1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -1 & 3 \\ 1 & -4 \end{vmatrix} = (-1)(-4) - (3)(1) = 4 - 3 = 1\)

2. \(\begin{vmatrix} 2 & 3 \\ 1 & -4 \end{vmatrix} = (2)(-4) - (3)(1) = -8 - 3 = -11\)

3. \(\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (-1)(1) = 2 + 1 = 3\)

Now substituting back:

\[ \vec{a} \times \vec{b} = \hat{i}(1) - \hat{j}(-11) + \hat{k}(3) = \hat{i} + 11\hat{j} + 3\hat{k} \]

### Step 3: Use the Given Relationship
We have the relationship:

\[ (\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c}) \]

Using the vector triple product identity, we can rewrite both sides:

1. Left Side: \((\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}\)

2. Right Side: \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}\)

Setting both expressions equal gives:

\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

### Step 4: Simplifying the Equation
From the equation, we can cancel \((\vec{a} \cdot \vec{c}) \vec{b}\) from both sides:

\[ -(\vec{b} \cdot \vec{c}) \vec{a} = -(\vec{a} \cdot \vec{b}) \vec{c} \]

This simplifies to:

\[ \vec{b} \cdot \vec{c} \vec{a} = \vec{a} \cdot \vec{b} \vec{c} \]

### Step 5: Conclude the Relationship
This implies that \(\vec{c}\) is a scalar multiple of \(\vec{a}\):

\[ \vec{c} = \lambda \vec{a} \]

for some scalar \(\lambda\).

### Step 6: Identify the Correct Option
Given the options, we need to find which vector is a scalar multiple of \(\vec{a}\). Since \(\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}\), we can check the options to find which one is a multiple of \(\vec{a}\).