Question

The number of vectors of unit length perpendicular to the vectors $\overrightarrow{a}=2\hat{i}+\hat{j}+2\hat{k}\mathrm{and}\overrightarrow{b}=\hat{j}+\hat{k}\mathrm{is}$
(a) one
(b) two
(c) three
(d) infinite

Answer 1

$$\begin{gathered} \mathrm{Unit}\mathrm{vector}\mathrm{perpendicular}\mathrm{to}\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{is}\pm \frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|} \\ \therefore \mathrm{for}\overrightarrow{a}=2\hat{i}+\hat{j}+2\hat{k}\mathrm{and}\overrightarrow{b}=\hat{j}+\hat{k} \\ \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 2\\ 0& 1& 1\end{array}\right| \\ \overrightarrow{a}\times \overrightarrow{b}=\hat{i}\left(1-2\right)-\hat{j}\left(2-0\right)+\hat{k}\left(2-0\right) \end{gathered}$$

$\begin{array}{rcl}\therefore \left|\overrightarrow{a}\times \overrightarrow{b}\right|& =& \sqrt{{\left(-1\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2}\\ & =& \sqrt{9}=3\end{array}$
$$\begin{gathered} \therefore \mathrm{unit}\mathrm{vector}\mathrm{perpendicular}\mathrm{to}\mathrm{both}\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{is}\pm \frac{-\hat{i}-2\hat{j}+2\hat{k}}{3} \\ =\pm \left(-\frac{\hat{i}}{3}-\frac{2\hat{j}}{3}+\frac{2\hat{k}}{3}\right) \end{gathered}$$

i.e. number of vectors of unit length perpendicular to the vectors $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ is two.

Hence, the correct answer is option B.