Question

The number of vectors of unit length perpendicular to vectors $\overrightarrow{a}=\hat{i}+\hat{j}\mathrm{and}\overrightarrow{b}=\hat{j}+\hat{k}\mathrm{is}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

The unit vector perpendicular to $\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{is}$given by $\pm \frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}$

$\begin{array}{rcl}\mathrm{for}\overrightarrow{a}& =& \hat{i}+\hat{j}\mathrm{and}\overrightarrow{b}=\hat{j}+\hat{k}\\ \overrightarrow{a}\times \overrightarrow{b}& =& \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|\\ & =& \hat{i}\left(1-0\right)-\hat{j}\left(1-0\right)+\hat{k}\left(1-0\right)\end{array}$
$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k} \\ \mathrm{i}.\mathrm{e}.\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{1+1+1}=\sqrt{3} \\ \therefore \mathrm{number}\mathrm{of}\mathrm{unit}\mathrm{vectors}\mathrm{perpendicular}\mathrm{to}\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{is}\mathrm{two}\mathrm{i}.\mathrm{e}\pm \frac{\left(\hat{i}-\hat{j}+\hat{k}\right)}{\sqrt{3}} \end{gathered}$$