Question

Let $\overrightarrow{a}=-\hat{i}+\hat{j}$ and $\overrightarrow{b}=\hat{i}+3\hat{j}.$ Then angle between the vectors $4\overrightarrow{a}+\overrightarrow{b}$ and $\frac{1}{4}(7\overrightarrow{b}-\overrightarrow{a})$ is

• A. $30°$
• B. $45°$
• C. $60°$
• D. $90°$

Answer 1

The correct option is B $45°$

Assume
$\overrightarrow{x}=4\overrightarrow{a}+\overrightarrow{b}=4(-\hat{i}+\hat{j})+(\hat{i}+3\hat{j})=-3\hat{i}+7\hat{j}$
$\overrightarrow{y}=\frac{1}{4}(7\overrightarrow{b}-\overrightarrow{a})=\frac{1}{4}\left(7\right(\hat{i}+3\hat{j})-(-\hat{i}+\hat{j}\left)\right)=2\hat{i}+5\hat{j}$
Angle between two vectors will be
$\cos \theta =\frac{\overrightarrow{x}\cdot \overrightarrow{y}}{|\overrightarrow{x}||\overrightarrow{y}|}=\frac{(-3\hat{i}+7\hat{j})\cdot (2\hat{i}+5\hat{j})}{|-3\hat{i}+7\hat{j}||2\hat{i}+5\hat{j}|}$
$\Rightarrow \cos \theta =\frac{(-6+35)}{\sqrt{3^2+7^2}\cdot \sqrt{2^2+5^2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =45°$