Question

If $\overrightarrow{a}+\overrightarrow{b}=2\overrightarrow{i}$ and $2\overrightarrow{a}-\overrightarrow{b}=\overrightarrow{i}-\overrightarrow{j}$ then cosine of the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

• A. ${\sin }^{-1}\frac{4}{5}$
• B. ${\cos }^{-1}\frac{4}{5}$
• C. ${\cos }^{-1}\frac{3}{5}$
• D. None of these

Answer 1

The correct option is B ${\cos }^{-1}\frac{4}{5}$

Given

$\overrightarrow{a}+\overrightarrow{b}=2\hat{i}$------------(1)

$2\overrightarrow{a}-\overrightarrow{b}=\hat{i}-\hat{j}$---------------(2)

Adding eq (1) and (2)

$\overrightarrow{a}+\overrightarrow{b}+2\overrightarrow{a}-\overrightarrow{b}=2\hat{i}+\hat{i}-\hat{j}$

$3\overrightarrow{a}=3\hat{i}-\hat{j}$

$\overrightarrow{a}=\hat{i}-\frac{1}{3}\hat{j}$

putting $\overrightarrow{a}$ in eq (1)

$\hat{i}-\frac{1}{3}\hat{j}+\overrightarrow{b}=2\hat{i}$

$\overrightarrow{b}=2\hat{i}-\hat{i}+\frac{1}{3}\hat{j}$

$\overrightarrow{b}=\hat{i}+\frac{1}{3}\hat{j}$

$\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$

$1-\frac{1}{9}=\sqrt{1^2+(\frac{1}{3}{)}^2}\sqrt{1^2+(\frac{1}{3}{)}^2}\cos \theta$

$\frac{8}{9}=\sqrt{\left(\frac{10}{9}\right)}\sqrt{\left(\frac{10}{9}\right)}\cos \theta$\

$\frac{8}{9}=\frac{10}{9}\cos \theta$

$\frac{4}{5}=\cos \theta$

$\theta =\cos -1\left(\frac{4}{5}\right)$