If $\hat a$ is a unit vector perpendicular to both the vectors $\vec b=\hat j+2\hat k$ and $\vec c= \hat i+ 2\hat j$, then $\hat a$ is equal to

$\frac{-4\hat i+2\hat j-\hat k}{\sqrt{21}}$

The correct answer is Option (3) → $\frac{-4\hat i+2\hat j-\hat k}{\sqrt{21}}$

Given:

$\vec b = \; \hat j + 2\hat k$

$\vec c = \; \hat i + 2\hat j$

A unit vector $\hat a$ perpendicular to both $\vec b$ and $\vec c$ is parallel to $\vec b \times \vec c$.

Compute the cross product:

$\vec b \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}$

$= \hat i(1\cdot 0 - 2\cdot 2)\;-\;\hat j(0\cdot 0 - 2\cdot 1)\;+\;\hat k(0\cdot 2 - 1\cdot 1)$

$= -4\hat i \;+\;2\hat j \;-\;\hat k$

Magnitude:

$|\vec b \times \vec c| = \sqrt{16 + 4 + 1} = \sqrt{21}$

Unit vector:

$\hat a = \frac{-4\hat i + 2\hat j - \hat k}{\sqrt{21}}$

Final Answer:

$\hat a = \frac{-4\hat i + 2\hat j - \hat k}{\sqrt{21}}$