If `veca=hati+hatj+hatk , vecb=2hati+hatj-hatk and vecc=4hati+3hatj+hatk` then value of `((veca+vecb)xx(veca -(veca-vecb)xxvecb)))xxvecc` is

Options

A. `30hati-34hatj+36hatk`
B. `30hati+34hatj+36hatk`
C. `30hati+34hatj-36hatk`
D. None of these

Answer

To solve the given problem, we will follow these steps:

1. **Define the vectors**:
- \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\)
- \(\vec{b} = 2\hat{i} + \hat{j} - \hat{k}\)
- \(\vec{c} = 4\hat{i} + 3\hat{j} + \hat{k}\)

2. **Calculate \(\vec{a} + \vec{b}\)**:
\[ \vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{j} - \hat{k}) \]
\[ = (1 + 2)\hat{i} + (1 + 1)\hat{j} + (1 - 1)\hat{k} = 3\hat{i} + 2\hat{j} + 0\hat{k} = 3\hat{i} + 2\hat{j} \]

3. **Calculate \(\vec{a} - \vec{b}\)**:
\[ \vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + \hat{j} - \hat{k}) \]
\[ = (1 - 2)\hat{i} + (1 - 1)\hat{j} + (1 + 1)\hat{k} = -\hat{i} + 0\hat{j} + 2\hat{k} = -\hat{i} + 2\hat{k} \]

4. **Calculate \((\vec{a} - \vec{b}) \times \vec{b}\)**:
\[ \vec{u} = \vec{a} - \vec{b} = -\hat{i} + 2\hat{k} \]
\[ \vec{u} \times \vec{b} = (-\hat{i} + 2\hat{k}) \times (2\hat{i} + \hat{j} - \hat{k}) \]
Using the determinant method:
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 0 & 2 \\ 2 & 1 & -1 \end{vmatrix} \]
\[ = \hat{i}(0 \cdot -1 - 2 \cdot 1) - \hat{j}(-1 \cdot -1 - 2 \cdot 2) + \hat{k}(-1 \cdot 1 - 0 \cdot 2) \]
\[ = \hat{i}(0 - 2) - \hat{j}(1 - 4) + \hat{k}(-1) \]
\[ = -2\hat{i} + 3\hat{j} - \hat{k} \]

5. **Calculate \((\vec{a} + \vec{b}) \times ((\vec{a} - \vec{b}) \times \vec{b})\)**:
\[ \vec{v} = \vec{a} + \vec{b} = 3\hat{i} + 2\hat{j} \]
\[ \vec{v} \times (-2\hat{i} + 3\hat{j} - \hat{k}) = (3\hat{i} + 2\hat{j}) \times (-2\hat{i} + 3\hat{j} - \hat{k}) \]
Using the determinant method:
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ -2 & 3 & -1 \end{vmatrix} \]
\[ = \hat{i}(2 \cdot -1 - 0 \cdot 3) - \hat{j}(3 \cdot -1 - 0 \cdot -2) + \hat{k}(3 \cdot 3 - 2 \cdot -2) \]
\[ = -2\hat{i} + 3\hat{j} + (9 + 4)\hat{k} \]
\[ = -2\hat{i} + 3\hat{j} + 13\hat{k} \]

6. **Calculate the final cross product with \(\vec{c}\)**:
\[ (-2\hat{i} + 3\hat{j} + 13\hat{k}) \times (4\hat{i} + 3\hat{j} + \hat{k}) \]
Using the determinant method:
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 13 \\ 4 & 3 & 1 \end{vmatrix} \]
\[ = \hat{i}(3 \cdot 1 - 13 \cdot 3) - \hat{j}(-2 \cdot 1 - 13 \cdot 4) + \hat{k}(-2 \cdot 3 - 3 \cdot 4) \]
\[ = \hat{i}(3 - 39) - \hat{j}(-2 - 52) + \hat{k}(-6 - 12) \]
\[ = -36\hat{i} + 54\hat{j} - 18\hat{k} \]

7. **Final Result**:
The final value is:
\[ -36\hat{i} + 54\hat{j} - 18\hat{k} \]