Question

The vectors $\overline{X}$ and $\overline{Y}$ satisfy the equations $2\overline{X}+\overline{Y}=\overline{p},\overline{X}+2\overline{Y}=\overline{q}$ where $\overline{p}=\overline{i}+\overline{j}$ and,$\overline{q}=\overline{i}-\overline{j}$. iF $\theta$ is the angle between $\overline{X}$ and $\overline{Y}$ then

• A. $\cos \theta =\frac{4}{5}$
• B. $\sin \theta =\frac{1}{\sqrt{2}}$
• C. $\cos \theta =-\frac{4}{5}$
• D. $\cos \theta =-\frac{3}{5}$

Answer 1

The correct option is C $\cos \theta =-\frac{4}{5}$

$\begin{array}{c}2\overline{X}+\overline{Y}=\overline{p}-----\left(1\right)\\ \\ \overline{X}+2\overline{Y}=\overline{q}-----\left(2\right)\end{array}$

Subtracting $2\times \left(2\right)$ from $\left(1\right)$, we get

$\begin{array}{c}2\overline{X}+\overline{Y}-\left(2\overline{X}+4\overline{Y}\right)=\overline{p}-2\overline{q}\\ \\ -3\overline{Y}=\overline{p}-2\overline{q}\\ \\ -3\overline{Y}=\left(\hat{i}+\hat{j}\right)-2\left(\hat{i}-\hat{j}\right)\\ \\ \overline{Y}=\frac{1}{3}\hat{i}-\hat{j}\end{array}$

Putting value of $\overline{Y}$ in $\left(1\right)$

$\begin{array}{c}2\overline{X}+\overline{Y}=\overline{p}\\ \\ 2\overline{X}+\frac{1}{3}\hat{i}-\hat{j}=\hat{i}-\hat{j}\\ \\ 2\overline{X}=\frac{2}{3}\hat{i}+2\hat{j}\\ \\ \overline{X}=\frac{1}{3}\hat{i}+\hat{j}\end{array}$

$\begin{array}{c}2\overline{X}+\overline{Y}=\overline{p}\\ \\ 2\overline{X}+\frac{1}{3}\hat{i}-\hat{j}=\hat{i}-\hat{j}\\ \\ 2\overline{X}=\frac{2}{3}\hat{i}+2\hat{j}\\ \\ \overline{X}=\frac{1}{3}\hat{i}+\hat{j}\\ \\ \left|\overline{X}\right|=\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(1\right)}^2}=\frac{\sqrt{10}}{3}\\ \\ \left|\overline{Y}\right|=\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(-1\right)}^2}=\frac{\sqrt{10}}{3}\\ \\ \overline{X}\cdot \overline{Y}=\left|\overline{X}\right|\left|\overline{Y}\right|\cos \theta \\ \\ \frac{1}{3}.\frac{1}{3}+1\left(-1\right)=\frac{\sqrt{10}}{3}\times \frac{\sqrt{10}}{3}\cos \theta \\ \\ \frac{1}{2}-1=\frac{10}{9}\cos \theta \\ \\ \cos \theta =-\frac{4}{5}\end{array}$

Hence, option $C$ is correct.