Let `a=3hati + 2hatj + 2hatk` and `b= hati + 2hatj - 2hatk` be two vectors. If a vector perpendicular to both the `a + b` and `a-b` has the magnitude 12, then one such vector is:

Options

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

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and elaborate on each step.

### Step 1: Define the vectors
Given:
- \( \mathbf{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \)
- \( \mathbf{b} = \hat{i} + 2\hat{j} - 2\hat{k} \)

### Step 2: Calculate \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \)
First, we find \( \mathbf{a} + \mathbf{b} \):
\[ \mathbf{a} + \mathbf{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} - 2\hat{k}) = (3 + 1)\hat{i} + (2 + 2)\hat{j} + (2 - 2)\hat{k} = 4\hat{i} + 4\hat{j} + 0\hat{k} \]
Thus,
\[ \mathbf{a} + \mathbf{b} = 4\hat{i} + 4\hat{j} \]

Next, we calculate \( \mathbf{a} - \mathbf{b} \):
\[ \mathbf{a} - \mathbf{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} - 2\hat{k}) = (3 - 1)\hat{i} + (2 - 2)\hat{j} + (2 + 2)\hat{k} = 2\hat{i} + 0\hat{j} + 4\hat{k} \]
Thus,
\[ \mathbf{a} - \mathbf{b} = 2\hat{i} + 4\hat{k} \]

### Step 3: Find a vector perpendicular to both \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \)
Let \( \mathbf{C} \) be the required vector. Since \( \mathbf{C} \) is perpendicular to both \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \), we can express \( \mathbf{C} \) as:
\[ \mathbf{C} = \lambda ((\mathbf{a} + \mathbf{b}) \times (\mathbf{a} - \mathbf{b})) \]

### Step 4: Calculate the cross product
Using the determinant method for the cross product:
\[ \mathbf{C} = \lambda \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(4 \cdot 4 - 0 \cdot 0) - \hat{j}(4 \cdot 4 - 0 \cdot 2) + \hat{k}(4 \cdot 0 - 4 \cdot 2) \]
\[ = \hat{i}(16) - \hat{j}(16) - \hat{k}(8) \]
Thus,
\[ \mathbf{C} = \lambda (16\hat{i} - 16\hat{j} - 8\hat{k}) \]

### Step 5: Simplify \( \mathbf{C} \)
Factoring out 8:
\[ \mathbf{C} = 8\lambda (2\hat{i} - 2\hat{j} - \hat{k}) \]

### Step 6: Set the magnitude of \( \mathbf{C} \) to 12
The magnitude of \( \mathbf{C} \) is given by:
\[ |\mathbf{C}| = |8\lambda| \sqrt{(2^2) + (-2)^2 + (-1)^2} = |8\lambda| \sqrt{4 + 4 + 1} = |8\lambda| \sqrt{9} = |8\lambda| \cdot 3 \]
Setting this equal to 12:
\[ |8\lambda| \cdot 3 = 12 \]
\[ |8\lambda| = 4 \quad \Rightarrow \quad |\lambda| = \frac{4}{8} = \frac{1}{2} \]

### Step 7: Substitute \( \lambda \) back into \( \mathbf{C} \)
Assuming \( \lambda = \frac{1}{2} \):
\[ \mathbf{C} = 8 \cdot \frac{1}{2} (2\hat{i} - 2\hat{j} - \hat{k}) = 4(2\hat{i} - 2\hat{j} - \hat{k}) = 8\hat{i} - 8\hat{j} - 4\hat{k} \]

Thus, one such vector \( \mathbf{C} \) is:
\[ \mathbf{C} = 8\hat{i} - 8\hat{j} - 4\hat{k} \]