A unit vector perpendicular to both the vectors \( 4\hat{\imath}-\hat{\jmath}+3\hat{k} \) and \( -2\hat{\imath}+\hat{\jmath}-2\hat{k} \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{3}\left(-\hat{\imath}+2\hat{\jmath}+2\hat{k}\right) \)
The cross product \( \vec{a}\times\vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \). Let \( \vec{a}=4\hat{\imath}-\hat{\jmath}+3\hat{k} \) and \( \vec{b}=-2\hat{\imath}+\hat{\jmath}-2\hat{k} \).
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\4 & -1 & 3\\-2 & 1 & -2\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(2-3)-\hat{\jmath}(-8+6)+\hat{k}(4-2) \)
\( \displaystyle =-\hat{\imath}+2\hat{\jmath}+2\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{1+4+4}=3 \)
\( \displaystyle \text{Unit vector}=\pm\frac{1}{3}\left(-\hat{\imath}+2\hat{\jmath}+2\hat{k}\right) \)
The positive sign gives Option (3).
Shortcut: All four options have magnitude 1, so only perpendicularity needs checking. Dot each bracket with \( 4\hat{\imath}-\hat{\jmath}+3\hat{k} \):
\( \displaystyle (1,2,2):\ 4-2+6=8 \)
\( \displaystyle (-1,2,2):\ -4-2+6=0 \)
\( \displaystyle (-1,-2,2):\ -4+2+6=4 \)
Only \( -\hat{\imath}+2\hat{\jmath}+2\hat{k} \) is perpendicular to the first vector, and it also gives \( 2+2-4=0 \) with the second.