Question

The cousines of the angles made by the vector $\hat{i}-2\hat{j}+2\hat{k}$ with the coordinate axes are: ______________.

Answer 1

Given: $\overrightarrow{a}=\hat{i}-2\hat{j}+2\hat{k}$


Let l, m and n are the direction cosines of the given vector, then
$$\begin{gathered} l=\frac{a_1}{\overrightarrow{\left|a\right|}} \\ =\frac{1}{\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(2\right)}^2}} \\ =\frac{1}{\sqrt{1+4+4}} \\ =\frac{1}{\sqrt{9}} \\ =\frac{1}{3} \\ m=\frac{a_2}{\overrightarrow{\left|a\right|}} \\ =\frac{-2}{\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(2\right)}^2}} \\ =\frac{-2}{\sqrt{1+4+4}} \\ =\frac{-2}{\sqrt{9}} \\ =-\frac{2}{3} \\ n=\frac{a_3}{\overrightarrow{\left|a\right|}} \\ =\frac{2}{\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(2\right)}^2}} \\ =\frac{2}{\sqrt{1+4+4}} \\ =\frac{2}{\sqrt{9}} \\ =\frac{2}{3} \\ \mathrm{Thus},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{are}\left(\frac{1}{3},-\frac{2}{3},\frac{2}{3}\right). \end{gathered}$$


Hence, the cosines of the angles made by the vector $\hat{i}-2\hat{j}+2\hat{k}$ with the coordinate axes are $\underline{\left(\frac{1}{3},-\frac{2}{3},\frac{2}{3}\right)}.$