Question

Find the angle between the vectors $\overrightarrow{a}=\hat{i}-\hat{j}+\hat{k}\mathrm{and}\overrightarrow{b}=\hat{i}+\hat{j}-\hat{k}.$

Answer 1

$$\begin{gathered} \text{We have} \\ \overrightarrow{a}=\hat{i}-\hat{j}+\hat{k}\mathrm{and}\overrightarrow{b}=i+j-k \\ \text{Let}\theta \text{be the angle between}\overrightarrow{a}\text{and}\overrightarrow{b}. \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2+{\left(1\right)}^2}=\sqrt{3} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(-1\right)}^2}=\sqrt{3} \\ \mathrm{and} \\ \overrightarrow{a}.\overrightarrow{b}=1-1-1=-1 \\ \mathrm{Now}, \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{-1}{\sqrt{3}\sqrt{3}}=\frac{-1}{3} \\ \therefore \theta ={\cos }^{-1}\left(\frac{-1}{3}\right) \end{gathered}$$