Discussion
Unit vector perpendicular to \(\bar{A} = 2\hat{i} - \hat{j} + \hat{k}\) and \(\bar{B} = 3\hat{i} + 4\hat{j} - \hat{k}\) is :
Answer: A
The unit vector is \(\frac{-3\hat{i} + 5\hat{j} + 11\hat{k}}{\sqrt{155}}\).
A vector perpendicular to both is their cross product:
\[ \begin{aligned} \bar{A} \times \bar{B} &= \hat{i}[(-1)(-1) - (1)(4)] \\ &\quad - \hat{j}[(2)(-1) - (1)(3)] \\ &\quad + \hat{k}[(2)(4) - (-1)(3)] \\ &= -3\hat{i} + 5\hat{j} + 11\hat{k} \end{aligned} \]
Its magnitude is \(\sqrt{9 + 25 + 121} = \sqrt{155}\), and dividing by it gives the unit vector. A quick check: its dot product with \(\bar{A}\) is \(-6 - 5 + 11 = 0\).
The other options fail that check; for example \((\hat{i} - \hat{j} + 2\hat{k})\cdot\bar{A} = 2 + 1 + 2 = 5\), so it is not perpendicular to \(\bar{A}\).
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