Question

Find the sine of the angle between the vectors $\overrightarrow{a}=3\hat{i}+\hat{j}+2\hat{k}$ and $\overrightarrow{b}=2\hat{i}-2\hat{j}+4\hat{k}$

Answer 1

Given, $\overrightarrow{a}=3\hat{i}+\hat{j}+2\hat{k}$ and $\overrightarrow{b}=2\hat{ı}-2\hat{ȷ}+4\hat{k}$ Angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$
Let $\theta$ is angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$

$\therefore \cos \theta =\frac{(3\hat{i}+\hat{j}+2\hat{k})\cdot (2\hat{i}-2\hat{j}+4\hat{k})}{\sqrt{3^2+1^2+2^2}\cdot \sqrt{2^2+2^2+4^2}}$

[From eq. (1) and (2)]

$=\frac{6-2+8}{\sqrt{14}\sqrt{24}}$

$=\frac{12}{4\sqrt{21}}\Rightarrow \cos \theta =\frac{3}{\sqrt{21}}\therefore \sin \theta =\sqrt{1-{\cos }^2\theta }[\because {\sin }^2\theta +{\cos }^2\theta =1]=\sqrt{1-\frac{9}{21}}\Rightarrow \sin \theta =\frac{2}{\sqrt{7}}$
Hence, sine of angle between vectors is $\frac{2}{\sqrt{7}}$