A vector of magnitude 9, which is perpendicular to both the vectors \((4\hat{\imath} -\hat{\jmath} +8\hat{k})\) and \((-\hat{\jmath}+\hat{k})\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(7\hat{\imath}-4\hat{\jmath}-4\hat{k}\)
Let \(\vec{a} = 4\hat{\imath} - \hat{\jmath} + 8\hat{k}\) and \(\vec{b} = -\hat{\jmath} + \hat{k}\). The cross product \(\vec{a} \times \vec{b}\) is perpendicular to both vectors:
\(\displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4 & -1 & 8 \\ 0 & -1 & 1 \end{vmatrix}\)
\(= \hat{\imath}(-1 + 8) - \hat{\jmath}(4 - 0) + \hat{k}(-4 - 0)\)
\(= 7\hat{\imath} - 4\hat{\jmath} - 4\hat{k}\)
Its magnitude is
\(\displaystyle |\vec{a} \times \vec{b}| = \sqrt{49 + 16 + 16}\)
\(= \sqrt{81} = 9\)
So \(\vec{a} \times \vec{b}\) itself has magnitude 9, and \(7\hat{\imath} - 4\hat{\jmath} - 4\hat{k}\) is the required vector. (Its negative, \(-7\hat{\imath} + 4\hat{\jmath} + 4\hat{k}\), also works but is not among the options.)