For any vector $\mathbf{a}$, the value of $(\mathbf{a} \times \hat{i})^2 + (\mathbf{a} \times \hat{j})^2 + (\mathbf{a} \times \hat{k})^2$ is

$2a^2$

The correct answer is Option (4) → $2a^2$ ##

Let $\mathbf{a} = x\hat{i} + y\hat{j} + z\hat{k}$

$∴a^2 = x^2 + y^2 + z^2$ ...(i)

$∴\mathbf{a} \times \hat{i} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 0 & 0 \end{matrix} \right|$

$= \hat{i}[0] - \hat{j}[-z] + \hat{k}[-y]$

$= z\hat{j} - y\hat{k}$

$∴(\mathbf{a} \times \hat{i})^2 = (z\hat{j} - y\hat{k}) \cdot (z\hat{j} - y\hat{k})$

$= y^2 + z^2$

Similarly, $(\mathbf{a} \times \hat{j})^2 = x^2 + z^2$

and $(\mathbf{a} \times \hat{k})^2 = x^2 + y^2$

$∴(\mathbf{a} \times \hat{i})^2 + (\mathbf{a} \times \hat{j})^2 + (\mathbf{a} \times \hat{k})^2 = y^2 + z^2 + x^2 + z^2 + x^2 + y^2$

$= 2(x^2 + y^2 + z^2) = 2\mathbf{a}^2 \quad \text{[using ...(i)]}$