Let `a=3hat(i) +2hat(j) +2hat(k) " and " b=hat(i) +2hat(j) -2hat(k)` be two vectors. If a vectors perpendicular to both the vectors a+b and a-b has the megnitude 12 ,then one such vectors is

Options

A. `4(2hat(i)+2hat(j)+hat(k))`
B. `4(2hat(i)-2hat(j)-hat(k))`
C. `4(2hat(i)+2hat(j)-hat(k))`
D. `4(-2hat(i)-2hat(j)+hat(k))`

Explanation

To solve the problem, we need to find a vector that is perpendicular to both \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \) with a magnitude of 12. Let's go through the steps systematically.

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

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

### Step 3: Find the cross product \( \mathbf{c} = (\mathbf{a} + \mathbf{b}) \times (\mathbf{a} - \mathbf{b}) \)
We need to calculate the cross product:
\[ \mathbf{c} = (4\hat{i} + 4\hat{j}) \times (2\hat{i} + 4\hat{k}) \]
Using the determinant method:
\[ \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 4 & 0 \\ 0 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 0 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 4 \\ 2 & 0 \end{vmatrix} \]
\[ = \hat{i} (4 \cdot 4 - 0 \cdot 0) - \hat{j} (4 \cdot 4 - 0 \cdot 2) + \hat{k} (4 \cdot 0 - 4 \cdot 2) \]
\[ = 16\hat{i} - 16\hat{j} - 8\hat{k} \]

### Step 4: Simplify the vector \( \mathbf{c} \)
We can factor out 8:
\[ \mathbf{c} = 8(2\hat{i} - 2\hat{j} - \hat{k}) \]

### Step 5: Find the unit vector in the direction of \( \mathbf{c} \)
Now, we find the magnitude of \( \mathbf{c} \):
\[ |\mathbf{c}| = \sqrt{(16)^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24 \]
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{c} \) is:
\[ \mathbf{u} = \frac{\mathbf{c}}{|\mathbf{c}|} = \frac{8(2\hat{i} - 2\hat{j} - \hat{k})}{24} = \frac{1}{3}(2\hat{i} - 2\hat{j} - \hat{k}) = \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \]

### Step 6: Scale the unit vector to have a magnitude of 12
To find the vector with magnitude 12, we multiply the unit vector by 12:
\[ \mathbf{d} = 12 \cdot \mathbf{u} = 12 \left(\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}\right) = 8\hat{i} - 8\hat{j} - 4\hat{k} \]

### Final Answer
Thus, one such vector that is perpendicular to both \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \) with a magnitude of 12 is:
\[ \mathbf{d} = 8\hat{i} - 8\hat{j} - 4\hat{k} \]