Let $$\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$$ and $$\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$$ be two vectors. If a vector perpendicular to both the vectors $$\vec{a} + \vec{b}$$ and $$\vec{a} - \vec{b}$$ has the magnitude 12 then one such vector is:

Given the vectors:

$$\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$$ $$\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$$

First, we find the sum and difference of the vectors:

• Addition: $$\vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j} + 0\hat{k}$$
• Subtraction: $$\vec{a} - \vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 0\hat{j} + 4\hat{k}$$

A vector perpendicular to both $$(\vec{a} + \vec{b})$$ and $$(\vec{a} - \vec{b})$$ is given by their cross product. Let $$\vec{u} = \vec{a} + \vec{b}$$ and $$\vec{v} = \vec{a} - \vec{b}$$:

$$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}$$

Expanding the determinant:

$$\vec{u} \times \vec{v} = \hat{i}(16 - 0) - \hat{j}(16 - 0) + \hat{k}(0 - 8)$$ $$\vec{u} \times \vec{v} = 16\hat{i} - 16\hat{j} - 8\hat{k}$$

The magnitude of this cross product is:

$$|\vec{u} \times \vec{v}| = \sqrt{16^2 + (-16)^2 + (-8)^2}$$ $$|\vec{u} \times \vec{v}| = \sqrt{256 + 256 + 64} = \sqrt{576} = 24$$

The question requires a vector with a magnitude of 12. Since our current vector has a magnitude of 24, we multiply the vector by $$\frac{12}{24}$$ (or $$\frac{1}{2}$$):

$$\text{Required Vector} = \frac{1}{2}(16\hat{i} - 16\hat{j} - 8\hat{k}) = 8\hat{i} - 8\hat{j} - 4\hat{k}$$

Factoring out a 4 to match the options provided:

$$8\hat{i} - 8\hat{j} - 4\hat{k} = 4(2\hat{i} - 2\hat{j} - \hat{k})$$

This matches Option B.