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 condition provided. Let's denote the vectors as follows:

- \(\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}\)
- \(\vec{b} = \hat{i} + \hat{j} - 4\hat{k}\)
- \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\)

We need to find the vector \(\vec{c}\) such that:
\[ (\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c}) \]

### Step 1: Calculate \(\vec{a} \times \vec{b}\)

Using the determinant method for the cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 1 & 1 & -4 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 3 \\ 1 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \\ 1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} \]

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

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

### Step 2: Calculate \((\vec{a} \times \vec{b}) \times \vec{c}\)

Now we need to calculate:
\[ (\vec{a} \times \vec{b}) \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 11 & 3 \\ x & y & z \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 11 & 3 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 11 \\ x & y \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 11 & 3 \\ y & z \end{vmatrix} = 11z - 3y\)
2. \(\begin{vmatrix} 1 & 3 \\ x & z \end{vmatrix} = z - 3x\)
3. \(\begin{vmatrix} 1 & 11 \\ x & y \end{vmatrix} = y - 11x\)

Thus,
\[ (\vec{a} \times \vec{b}) \times \vec{c} = (11z - 3y)\hat{i} - (z - 3x)\hat{j} + (y - 11x)\hat{k} \]

### Step 3: Calculate \(\vec{b} \times \vec{c}\)

Now we calculate:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -4 \\ x & y & z \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & -4 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -4 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 1 & -4 \\ y & z \end{vmatrix} = z + 4y\)
2. \(\begin{vmatrix} 1 & -4 \\ x & z \end{vmatrix} = z + 4x\)
3. \(\begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} = y - x\)

Thus,
\[ \vec{b} \times \vec{c} = (z + 4y)\hat{i} - (z + 4x)\hat{j} + (y - x)\hat{k} \]

### Step 4: Calculate \(\vec{a} \times (\vec{b} \times \vec{c})\)

Now we calculate:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ (z + 4y) & -(z + 4x) & (y - x) \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 3 \\ -(z + 4x) & (y - x) \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \\ (z + 4y) & (y - x) \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ (z + 4y) & -(z + 4x) \end{vmatrix} \]

Calculating these determinants will yield the final expression for \(\vec{a} \times (\vec{b} \times \vec{c})\).

### Step 5: Set the two expressions equal

Now we equate the coefficients from both sides of the equation:
\[ (\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c}) \]

### Step 6: Solve for \(x\), \(y\), and \(z\)

From the equations obtained, we can derive relationships between \(x\), \(y\), and \(z\).

### Final Step: Express \(\vec{c}\)

Once we have the relationships, we can express \(\vec{c}\) in terms of a parameter \(m\) (if necessary) to find the vector \(\vec{c}\).