A vector of magnitude 8 units in the direction perpendicular to both the vectors \(\hat{\imath}+\hat{\jmath}+\hat{k}\) and \(2\hat{\imath}+\hat{k}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{8}{\sqrt{6}}(\hat{\imath}+\hat{\jmath}-2\hat{k})\)
Let \(\vec{a} = \hat{\imath} + \hat{\jmath} + \hat{k}\) and \(\vec{b} = 2\hat{\imath} + \hat{k}\). The cross product \(\vec{a} \times \vec{b}\) is perpendicular to both:
\( \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\1&1&1\\2&0&1\end{vmatrix} \)
\(= \hat{\imath}(1 - 0) - \hat{\jmath}(1 - 2) + \hat{k}(0 - 2)\)
\(= \hat{\imath} + \hat{\jmath} - 2\hat{k}\)
\(|\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 4} = \sqrt{6}\)
A vector of magnitude 8 along it is 8 times the unit vector:
\( \displaystyle 8 \cdot \frac{\hat{\imath} + \hat{\jmath} - 2\hat{k}}{\sqrt{6}} = \frac{8}{\sqrt{6}}(\hat{\imath} + \hat{\jmath} - 2\hat{k}) \)
Why the other options are incorrect
Option (1) \(8(\hat{\imath}+\hat{\jmath}-2\hat{k})\) — Incorrect. The direction is right, but \(\hat{\imath} + \hat{\jmath} - 2\hat{k}\) is not a unit vector, so the magnitude is \(8\sqrt{6}\), not 8.
Option (2) \(\displaystyle\frac{8}{\sqrt{6}}(-\hat{\imath}+\hat{\jmath}+2\hat{k})\) — Incorrect. Its dot product with \(\vec{a}\) is \(-1 + 1 + 2 = 2 \ne 0\), so it is not perpendicular to \(\vec{a}\).
Option (4) \(\displaystyle\frac{8}{\sqrt{6}}(\hat{\imath}-\hat{\jmath}+2\hat{k})\) — Incorrect. Its dot product with \(\vec{a}\) is \(1 - 1 + 2 = 2 \ne 0\), so it is not perpendicular to \(\vec{a}\).