Question

Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.

Answer 1

$$\begin{gathered} \mathrm{Let}\overrightarrow{\mathrm{a}}\mathrm{be}\mathrm{a}\mathrm{vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{vector}\mathrm{with}\mathrm{direction}\mathrm{ratios}2,3,6. \\ \Rightarrow \overrightarrow{a}=2\hat{i}+3\hat{j}+6\hat{k}. \\ \mathrm{Let}\overrightarrow{b}\mathrm{be}\mathrm{a}\mathrm{vector}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{vector}\mathrm{with}\mathrm{direction}\mathrm{ratios}1,2,2. \\ \Rightarrow \overrightarrow{b}=\hat{i}+2\hat{j}+2\hat{k} \\ \mathrm{Let}\theta \mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{the}\mathrm{given}\mathrm{vectors}. \\ \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|} \\ =\frac{\left(2\hat{i}+3\hat{j}+6\hat{k}\right).\left(\hat{i}+2\hat{j}+2\hat{k}\right)}{\left|2\hat{i}+3\hat{j}+6\hat{k}\right|\left|\hat{i}+2\hat{j}+2\hat{k}\right|} \\ =\frac{2+6+12}{\sqrt{4+9+36}\sqrt{1+4+4}} \\ =\frac{20}{21} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{20}{21}\right) \end{gathered}$$