Question

Find the cosine of the angle between the vectors $4\hat{i}-3\hat{j}+3\hat{k}\mathrm{and}2\hat{i}-\hat{j}-\hat{k}.$

Answer 1

$$\begin{gathered} \mathrm{Let},\overrightarrow{a}=4\hat{i}-3\hat{j}+3\hat{k} \\ \mathrm{and}\overrightarrow{b}=2\hat{i}-\hat{j}-\hat{k} \\ \text{Let}\theta \text{be the angle between}\overrightarrow{a}\text{and}\overrightarrow{b}. \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(4\right)}^2+{\left(-3\right)}^2+{\left(3\right)}^2}=\sqrt{34} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2+{\left(-1\right)}^2}=\sqrt{6} \\ \therefore \overrightarrow{a}.\overrightarrow{b}=8+3-3=8 \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{8}{\sqrt{34}\sqrt{6}}=\frac{8}{2\sqrt{51}}=\frac{4}{\sqrt{51}} \end{gathered}$$