Question

Determine unit vector which is perpendicular to both $\overrightarrow{A}=2\hat{i}+\hat{j}+\hat{k}\mathrm{and}\overrightarrow{B}=\hat{i}-\hat{j}+2\hat{k}$.

Answer 1

Dear Student,

$$\begin{gathered} \overrightarrow{\mathrm{A}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{B}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\left[\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 1& 1\\ 1& -1& 2\end{array}\right] \\ \overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\hat{\mathrm{i}}-3\hat{\mathrm{j}}+-3\hat{\mathrm{k}} \\ \left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|=\sqrt{{\left(1\right)}^2+{\left(-3\right)}^2+{\left(-3\right)}^2} \\ \left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|=\sqrt{19} \\ \mathrm{unit}\mathrm{vector}\mathrm{along}\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\frac{\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}}{\left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|} \\ \mathrm{unit}\mathrm{vector}\mathrm{along}\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\frac{\hat{\mathrm{i}}-3\hat{\mathrm{j}}+-3\hat{\mathrm{k}}}{\sqrt{19}} \end{gathered}$$