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

$\text{Here}{a}_1=3,a_2=1,a_3=2\text{and}{b}_1=2,b_2=-2,b_3=4\text{We know that,}cos\theta =\frac{a_1{b}_1+a_2{b}_2+a_3{b}_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}=\frac{3\times 2+1\times (-2)+2\times 4}{\sqrt{3^2+1^2+2^2\sqrt{2^2+(-2{)}^2+4^2}}}=\frac{6-2+8}{\sqrt{14}\sqrt{24}}=\frac{12}{2\sqrt{14}\sqrt{6}}=\frac{6}{\sqrt{84}}=\frac{6}{2\sqrt{21}}=\frac{3}{\sqrt{21}}\therefore sin\theta =\sqrt{1-co{s}^2\theta }=\sqrt{1-\frac{9}{21}}=\sqrt{\frac{12}{21}}=\frac{2\sqrt{3}}{\sqrt{3}\sqrt{7}}=\frac{2}{\sqrt{7}}$