Question

The vectors $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$satisfy the equations $2\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{p}\mathrm{and}\overrightarrow{a}+2\overrightarrow{b}=\overrightarrow{q},\mathrm{where}\overrightarrow{p}=\hat{i}+\hat{j}\mathrm{and}\overrightarrow{q}=\hat{i}-\hat{j}.$ If θ is the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$, then
(a) $\cos \mathrm{\theta }=\frac{4}{5}$

(b) $\sin \mathrm{\theta }=\frac{1}{\sqrt{2}}$

(c) $\cos \mathrm{\theta }=-\frac{4}{5}$

(d) $\cos \mathrm{\theta }=-\frac{3}{5}$

Answer 1

(c) $\cos \mathrm{\theta }=-\frac{4}{5}$

$$\begin{gathered} \text{Given that} \\ 2\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{p}...\left(1\right) \\ \overrightarrow{a}+2\overrightarrow{b}=\overrightarrow{q}...\left(2\right) \\ \text{Solving these two we get} \\ \overrightarrow{a}=\frac{2\overrightarrow{p}-\overrightarrow{q}}{3},\overrightarrow{b}=\frac{2\overrightarrow{q}-\overrightarrow{p}}{3} \\ \mathrm{And}\mathrm{we}\mathrm{have} \\ \overrightarrow{p}=\hat{i}+\hat{j}\mathrm{and}\overrightarrow{q}=\hat{i}-\hat{j} \\ \text{Substituting the values of}\overrightarrow{p}\text{and}\overrightarrow{q,}\text{we get} \\ \overrightarrow{a}=\frac{2\overrightarrow{p}-\overrightarrow{q}}{3}=\frac{2\left(\hat{i}+\hat{j}\right)-\left(\hat{i}-\hat{j}\right)}{3}=\frac{\hat{i}+3\hat{j}}{3} \\ \Rightarrow \left|\overrightarrow{a}\right|=\frac{1}{3}\sqrt{1+9}=\frac{\sqrt{10}}{3} \\ \overrightarrow{b}=\frac{2\overrightarrow{q}-\overrightarrow{p}}{3}=\frac{2\left(\hat{i}-\hat{j}\right)-\left(\hat{i}+\hat{j}\right)}{3}=\frac{\hat{i}-3\hat{j}}{3} \\ \Rightarrow \left|\overrightarrow{b}\right|=\frac{1}{3}\sqrt{1+9}=\frac{\sqrt{10}}{3} \\ \overrightarrow{a}.\overrightarrow{b}=\frac{1}{9}\left(1-9\right)=\frac{-8}{9} \\ \text{We know that} \\ \overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ \Rightarrow \frac{-8}{9}=\frac{\sqrt{10}}{3}\times \frac{\sqrt{10}}{3}\cos \theta \\ \Rightarrow \frac{-8}{9}=\frac{10}{9}\cos \theta \\ \Rightarrow \cos \theta =\frac{-8}{9}\times \frac{9}{10}=\frac{-4}{5} \end{gathered}$$