A) \[\frac{1}{3}\,(\mathbf{i}-2\mathbf{j}+2\mathbf{k})\]
B) \[\frac{1}{3}\,(-\mathbf{i}+2\mathbf{j}+2\mathbf{k})\]
C) \[\frac{1}{3}\,(2\mathbf{i}+\mathbf{j}+2\mathbf{k})\]
D) \[\frac{1}{3}\,(2\mathbf{i}-2\mathbf{j}+2\mathbf{k})\]
Correct Answer: B
Solution :
Let \[\mathbf{a}=4\mathbf{i}-\mathbf{j}+3\mathbf{k}\] and \[\mathbf{b}=-2\mathbf{i}+\mathbf{j}-2\mathbf{k}\] Unit vector perpendicular to \[\mathbf{a}\] and \[\mathbf{b}\] is \[\frac{\mathbf{a}\times \mathbf{b}}{|\mathbf{a}\times \mathbf{b}|}\] But \[\mathbf{a}\times \mathbf{b}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \\ \end{matrix} \right|\] \[=\mathbf{i}(2-3)-\mathbf{j}(-8+6)+\mathbf{k}(4-2)=-\mathbf{i}+2\mathbf{j}+2\mathbf{k}\] \[\therefore \,\frac{\mathbf{a}\times \mathbf{b}}{|\mathbf{a}\times \mathbf{b}|}=\frac{-\mathbf{i}+2\mathbf{j}+2\mathbf{k}}{\sqrt{1+4+4}}=\frac{-\mathbf{i}+2\mathbf{j}+2\mathbf{k}}{3}.\] Trick : Check it with the options. Since the vector \[\frac{-\mathbf{i}+2\mathbf{j}+2\mathbf{k}}{3}\] is unit and perpendicular to both the given vectors.
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