Question

find a unit vector in the plane of vectors a=i+2j and b=j+2k and perpendicular to the vector c= 2i+j+2k

Answer 1

Dear Student,

The vectors are :

$$\begin{gathered} \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+0\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{b}}=0\hat{\mathrm{i}}+\hat{\mathrm{j}}+2\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{c}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}+2\hat{\mathrm{k}} \end{gathered}$$

Consider,

$$\begin{gathered} \left(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right)=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 2& 0\\ 0& 1& 2\end{array}\right|=4\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \left(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right)\times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 4& -2& 1\\ 2& 1& 2\end{array}\right|=-5\hat{\mathrm{i}}-6\hat{\mathrm{j}}+8\hat{\mathrm{k}} \\ \left|\left(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right)\times \overrightarrow{\mathrm{c}}\right|=\sqrt{{\left(-5\right)}^2+{\left(-6\right)}^2+{\left(8\right)}^2}=\sqrt{25+36+64}=\sqrt{125}=5\sqrt{5} \\ \mathrm{So},\mathrm{required}\mathrm{unit}\mathrm{vector}=\frac{\left(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right)\times \overrightarrow{\mathrm{c}}}{\left|\left(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right)\times \overrightarrow{\mathrm{c}}\right|}=\frac{1}{5\sqrt{5}}\left[-5\hat{\mathrm{i}}-6\hat{\mathrm{j}}+8\hat{\mathrm{k}}\right] \end{gathered}$$

Regards