Question

Find the angle between the vectors whose direction cosines are proportional to 2, 3, −6 and 3, −4, 5.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathit{}\overrightarrow{a}\mathrm{be}\mathrm{a}\mathrm{vector}\mathrm{with}\mathrm{direction}\mathrm{ratios}2,3,-6. \\ \Rightarrow \overrightarrow{a}\mathit{}\mathit{=}\mathit{}\mathit{2}\hat{i}\mathit{}\mathit{+}\mathit{3}\hat{j}\mathit{}\mathit{-}\mathit{6}\hat{k}. \\ \mathrm{Let}\overrightarrow{b}\mathrm{be}\mathrm{a}\mathrm{vector}\mathrm{with}\mathrm{direction}\mathrm{ratios}3,-4,5. \\ \Rightarrow \overrightarrow{b}\mathit{}\mathit{=}\mathit{}\mathit{3}\hat{i}\mathit{}\mathit{-}\mathit{4}\hat{j}\mathit{}\mathit{+}\mathit{}\mathit{5}\hat{k} \\ \mathrm{Let}\theta \mathit{}\mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{given}\mathrm{vectors}. \\ \mathrm{Now}, \\ \cos \theta =\mathit{}\frac{\overrightarrow{a}\mathit{.}\mathit{}\overrightarrow{b}}{\left|\overrightarrow{a}\right|\mathit{}\left|\overrightarrow{b}\right|} \\ =\frac{\left(\mathit{2}\hat{i}\mathit{}\mathit{+}\mathit{3}\hat{j}\mathit{}\mathit{-}\mathit{6}\hat{k}\right).\left(\mathit{3}\hat{i}\mathit{}\mathit{-}\mathit{4}\hat{j}\mathit{}\mathit{+}\mathit{}\mathit{5}\hat{k}\right)}{\left|\mathit{2}\hat{i}\mathit{}\mathit{+}\mathit{3}\hat{j}\mathit{}\mathit{-}\mathit{6}\hat{k}\right|\left|\mathit{3}\hat{i}\mathit{}\mathit{-}\mathit{4}\hat{j}\mathit{}\mathit{+}\mathit{}\mathit{5}\hat{k}\right|} \\ =\frac{6-12-30}{\sqrt{4+9+36}\sqrt{9+16+25}} \\ =\frac{-36}{\sqrt{49}\sqrt{50}} \\ =\frac{-36}{35\sqrt{2}} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{result},\mathrm{we}\mathrm{get} \\ \cos \theta =-\frac{18\sqrt{2}}{35} \\ \therefore \theta ={\cos }^{-1}\left(-\frac{18\sqrt{2}}{35}\right) \\ \mathrm{Thus},\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{given}\mathrm{vectors}\mathrm{measures}{\cos }^{-1}\left(-\frac{18\sqrt{2}}{35}\right). \end{gathered}$$