A vector of magnitude 9, which is 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 1
The correct answer is Option (1) → \( -3\hat{\imath}+6\hat{\jmath}+6\hat{k} \)
A vector perpendicular to both \( \vec{a}=4\hat{\imath}-\hat{\jmath}+3\hat{k} \) and \( \vec{b}=-2\hat{\imath}+\hat{\jmath}-2\hat{k} \) is the cross product \( \vec{a}\times\vec{b} \):
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\4 & -1 & 3\\-2 & 1 & -2\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(2-3)-\hat{\jmath}(-8+6)+\hat{k}(4-2) \)
\( \displaystyle =-\hat{\imath}+2\hat{\jmath}+2\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{1+4+4}=3 \)
A vector of magnitude 9 in this direction is 9 times the unit vector:
\( \displaystyle 9\cdot\frac{-\hat{\imath}+2\hat{\jmath}+2\hat{k}}{3}=-3\hat{\imath}+6\hat{\jmath}+6\hat{k} \)
Shortcut: All four options have magnitude \( \sqrt{9+36+36}=9 \), so only perpendicularity needs checking. Dot each with \( 4\hat{\imath}-\hat{\jmath}+3\hat{k} \): Option (1) gives −12 − 6 + 18 = 0; Option (2) −12 + 6 + 18 = 12; Option (3) 12 + 6 + 18 = 36; Option (4) 12 − 6 − 18 = −12. Only Option (1) is perpendicular; it also gives 6 + 6 − 12 = 0 with the second vector.