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Question: A unit vector perpendicular to the vector \(4\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(- 2\mathb...

A unit vector perpendicular to the vector 4i−j+3k4\mathbf{i} - \mathbf{j} + 3\mathbf{k} and −2i+j−2k- 2\mathbf{i} + \mathbf{j} - 2\mathbf{k} is

A

13(i−2j+2k)\frac{1}{3}(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})

B

13(−i+2j+2k)\frac{1}{3}( - \mathbf{i} + 2\mathbf{j} + 2\mathbf{k})

C

13(2i+j+2k)\frac{1}{3}(2\mathbf{i} + \mathbf{j} + 2\mathbf{k})

D

13(2i−2j+2k)\frac{1}{3}(2\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})

Answer

13(−i+2j+2k)\frac{1}{3}( - \mathbf{i} + 2\mathbf{j} + 2\mathbf{k})

Explanation

Solution

Let a=4i−j+3k\mathbf{a} = 4\mathbf{i} - \mathbf{j} + 3\mathbf{k} and b=−2i+j−2k\mathbf{b} = - 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}

Unit vector perpendicular to a\mathbf{a} and b\mathbf{b} is a×b∣a×b∣\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|$

=i(2−3)−j(−8+6)+k(4−2)=−i+2j+2k= \mathbf{i}(2 - 3) - \mathbf{j}( - 8 + 6) + \mathbf{k}(4 - 2) = - \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}

∴a×b∣a×b∣=−i+2j+2k1+4+4=−i+2j+2k3.\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 −i+2j+2k3\frac{- \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{3} is unit and perpendicular to both the given

vectors.