Question

Find the angle between the vectors with direction ratios proportional to 1, −2, 1 and 4, 3, 2.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathit{}\overrightarrow{a}\mathrm{be}\mathrm{a}\mathrm{vector}\mathrm{with}\mathrm{direction}\mathrm{ratios}1,-2,1. \\ \Rightarrow \overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k}. \\ \mathrm{Let}\overrightarrow{b}\mathrm{be}\mathrm{a}\mathrm{vector}\mathrm{with}\mathrm{direction}\mathrm{ratios}4,3,2. \\ \Rightarrow \overrightarrow{b}\mathit{=}\mathit{4}\hat{i}\mathit{}\mathit{+}\mathit{}\mathit{3}\hat{j}\mathit{}\mathit{+}\mathit{}\mathit{2}\hat{k}. \\ \mathrm{Let}\theta \mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{given}\mathrm{vectors}. \\ \mathrm{Now}, \\ c\mathrm{os}\theta =\frac{\overrightarrow{a}\mathit{.}\overrightarrow{b}}{\left|\overrightarrow{a}\right|\mathit{}\left|\overrightarrow{b}\right|} \\ =\frac{\left(\hat{i}-2\hat{j}+\hat{k}\right).\left(\mathit{4}\hat{i}\mathit{}\mathit{+}\mathit{}\mathit{3}\hat{j}\mathit{}\mathit{+}\mathit{}\mathit{2}\hat{k}\right)}{\left|\hat{i}-2\hat{j}+\hat{k}\right|\left|\mathit{4}\hat{i}\mathit{}\mathit{+}\mathit{}\mathit{3}\hat{j}\mathit{}\mathit{+}\mathit{}\mathit{2}\hat{k}\right|} \\ =\frac{4-6+2}{\sqrt{1+4+1}\sqrt{16+9+4}} \\ =\frac{0}{\sqrt{6}\sqrt{29}} \\ =0 \\ \therefore \theta =\frac{\mathrm{\pi }}{2} \\ \mathrm{Thus},\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{given}\mathrm{vectors}\mathrm{measures}\frac{\mathrm{\pi }}{2}. \end{gathered}$$