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The number of vectors of unit length perpendicular to the two vectors $\mathbf{a}=(1,1,0)$ and $\mathbf{b}=(0,1,1)$ is
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The correct answer is:
$2$
Given, $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{b}=\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$,
$\pm \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}$
$\therefore$ Two vectors are possible
Unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$,
$\pm \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}$
$\therefore$ Two vectors are possible
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