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
Answer & explanation
Correct answer: option 3
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:
$\hat a = \frac{-4\hat i + 2\hat j - \hat k}{\sqrt{21}}$