Question

Let $\overrightarrow{a}=2\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}\mathrm{and}\overrightarrow{b}=3\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k}$. Find the angle between them.

Answer 1

We have:
$$\begin{gathered} \overrightarrow{a}=2\overrightarrow{\mathit{i}}+3\overrightarrow{\mathit{j}}+4\overrightarrow{k} \\ \overrightarrow{b}=3\overrightarrow{\mathit{i}}+4\overrightarrow{\mathit{j}}+5\overrightarrow{k} \end{gathered}$$

Using scalar product, we can find the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.
i.e., $\overrightarrow{\mathit{a}}\mathit{.}\overrightarrow{\mathit{b}}\mathit{=}\left|\overrightarrow{\mathit{a}}\right|\left|\overrightarrow{\mathit{b}}\right|\mathit{}\cos \theta$
So, $\theta ={\cos }^{-1}\left(\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)$

$$\begin{gathered} ={\cos }^{-1}\left(\frac{2\times 3+3\times 4+4\times 5}{\sqrt{\left(2^2+3^2+4^2\right)}\sqrt{\left(3^2+4^2+5^2\right)}}\right) \\ ={\cos }^{-1}\left(\frac{38}{\sqrt{29}\sqrt{50}}={\cos }^{-1}\frac{38}{\sqrt{1450}}\right) \end{gathered}$$

∴ The required angle is ${\cos }^{-1}\frac{38}{\sqrt{1450}}.$