The number of unit vectors perpendicular to $\overline{\mathrm{a}}=(1,1,0)$ and $\overline{\mathrm{b}}=(0,1,1)$ is

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Aicharya · Accepted Answer

The vector perpendicular to $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is $\bar{a} \times \bar{b}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 0 \end{array}\right|=\hat{i}-\hat{j}+\hat{k}$
Since the length of this vector is $\sqrt{3}$, the unit vector perpendicular to $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is $\pm \frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}= \pm \frac{1}{\sqrt{3}}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$ Hence, there are two such vectors.

The number of unit vectors perpendicular to $\overline{\mathrm{a}}=(1,1,0)$ and $\overline{\mathrm{b}}=(0,1…

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