Let \(\vec a = 3\hat{\imath}+\hat{\jmath}-4\hat{k}\) and \(\vec b= 6\hat{\imath}+5\hat{\jmath}-2\hat{k}\) be two vectors. Then a vector perpendicular to \(\vec a\) and \(\vec b\) with magnitude 3 units is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\hat{\imath}-2\hat{\jmath}+\hat{k}\)
A vector perpendicular to both \(\vec a\) and \(\vec b\) lies along the cross product:
\(\displaystyle \vec a \times \vec b = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\3&1&-4\\6&5&-2\end{vmatrix}\)
\(= (-2 + 20)\hat{\imath} - (-6 + 24)\hat{\jmath} + (15 - 6)\hat{k}\)
\(= 18\hat{\imath} - 18\hat{\jmath} + 9\hat{k}\)
\(|\vec a \times \vec b| = \sqrt{324 + 324 + 81} = \sqrt{729} = 27\)
The unit vector along it, multiplied by 3, has magnitude 3:
\(\displaystyle 3 \times \frac{18\hat{\imath} - 18\hat{\jmath} + 9\hat{k}}{27} = 2\hat{\imath} - 2\hat{\jmath} + \hat{k}\)
(Its negative, \(-2\hat{\imath} + 2\hat{\jmath} - \hat{k}\), also works, but it is not among the options.)
Shortcut: Options (3) and (4) have magnitude \(\frac{1}{3}\sqrt{4 + 4 + 1} = 1\), not 3. Of the other two, \(\vec a \cdot (2\hat{\imath} + 2\hat{\jmath} - \hat{k}) = 6 + 2 + 4 = 12 \ne 0\), while for Option (2):
\(\vec a \cdot (2\hat{\imath} - 2\hat{\jmath} + \hat{k}) = 6 - 2 - 4 = 0\)
\(\vec b \cdot (2\hat{\imath} - 2\hat{\jmath} + \hat{k}) = 12 - 10 - 2 = 0\)