A) Three
B) One
C) Two
D) Infinite
Correct Answer: C
Solution :
The vector perpendicular to \[\mathbf{a}\]and \[\mathbf{b}\] is \[\mathbf{a}\times \mathbf{b}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{matrix} \right|=\mathbf{i}-\mathbf{j}+\mathbf{k}\] Since the length of this vector is \[\sqrt{3},\]the unit vector perpendicular to \[\mathbf{a}\] and \[\mathbf{b}\] is \[\pm \frac{\mathbf{a}\times \mathbf{b}}{|\mathbf{a}\times \mathbf{b}|}=\pm \frac{1}{\sqrt{3}}(\mathbf{i}-\mathbf{j}+\mathbf{k})\] Hence there are two such vectors.
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