Question

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

(ii) $\overrightarrow{a}=3\hat{i}-2\hat{j}-6\hat{k}\mathrm{and}\overrightarrow{b}=4\hat{i}-\hat{j}+8\hat{k}$

(iii) $\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}\mathrm{and}\overrightarrow{b}=4\hat{i}+4\hat{j}-2\hat{k}$

(iv) $\overrightarrow{a}=2\hat{i}-3\hat{j}+\hat{k}\mathrm{and}\overrightarrow{b}=\hat{i}+\hat{j}-2\hat{k}$

(v) $\overrightarrow{a}=\hat{i}+2\hat{j}-\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$

Answer 1

$$\begin{gathered} \left(i\right)\text{Let}\theta \text{be the angle between}\overrightarrow{a}\text{and}\overrightarrow{b}. \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}=\sqrt{2} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{2} \\ \overrightarrow{a}.\overrightarrow{b}=0-1+0=-1 \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{-1}{\sqrt{2}\sqrt{2}}=\frac{-1}{2} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-1}{2}\right)=\frac{2\mathrm{\pi }}{3} \end{gathered}$$

$$\begin{gathered} \left(ii\right)\text{Let}\theta \text{be the angle between}\overrightarrow{a}\text{and}\overrightarrow{b}. \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(3\right)}^2+{\left(-2\right)}^2+{\left(-6\right)}^2}=\sqrt{49}=7 \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(4\right)}^2+{\left(-1\right)}^2+{\left(8\right)}^2}=\sqrt{81}=9 \\ \overrightarrow{a}.\overrightarrow{b}=12+2-48=-34 \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{-34}{\left(7\right)\left(9\right)}=\frac{-34}{63} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-34}{63}\right) \end{gathered}$$

$$\begin{gathered} \left(iii\right)\text{Let}\theta \text{be the angle between}\overrightarrow{a}\text{and}\overrightarrow{b}. \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2+{\left(2\right)}^2}=\sqrt{9}=3 \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(4\right)}^2+{\left(4\right)}^2+{\left(-2\right)}^2}=\sqrt{36}=6 \\ \overrightarrow{a}.\overrightarrow{b}=8-4-4=0 \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{0}{\left(3\right)\left(6\right)}=0 \\ \Rightarrow \theta ={\cos }^{-1}\left(0\right)=\frac{\mathrm{\pi }}{2} \end{gathered}$$

$$\begin{gathered} \left(iv\right)\text{Let}\theta \text{be the angle between}\overrightarrow{a}\text{and}\overrightarrow{b}. \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2+{\left(1\right)}^2}=\sqrt{14} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(-2\right)}^2}=\sqrt{6} \\ \overrightarrow{a}.\overrightarrow{b}=2-3-2=-3 \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{-3}{\sqrt{14}\sqrt{6}}=\frac{-3}{\sqrt{84}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-3}{\sqrt{84}}\right) \end{gathered}$$

$$\begin{gathered} \left(v\right)\text{Let}\theta \text{be the angle between}\overrightarrow{a}\text{and}\overrightarrow{b}. \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(-1\right)}^2}=\sqrt{6} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2+{\left(1\right)}^2}=\sqrt{3} \\ \overrightarrow{a}.\overrightarrow{b}=1-2-1=-2 \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{-2}{\sqrt{6}\sqrt{3}}=\frac{-2}{\sqrt{18}}=\frac{-\sqrt{2}\times \sqrt{2}}{\sqrt{2}\times \sqrt{9}}=\frac{-\sqrt{2}}{3} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-\sqrt{2}}{3}\right) \end{gathered}$$