A) one
B) two
C) three
D) infinite
Correct Answer: B
Solution :
We have \[\vec{a}=\hat{i}+\hat{j},\hat{b}=\hat{j}+\hat{k}\] Now, \[\vec{a}\times \vec{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{matrix} \right|\] \[=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(1-0)\] \[=\hat{i}-\hat{j}+\hat{k}\] \[\therefore \]Unit vectors \[=\pm \frac{\vec{a}\times \vec{b}}{|\vec{a}\times \vec{b}|}\] \[=\pm \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{1+1+1}}=\pm \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}\] So, there are two unit length perpendicular to the two vectors.
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