Question

Find the sine of the angle between the vectors $\hat{i}+2\hat{j}+2\hat{k}$ and $3\hat{i}+2\hat{j}+6\hat{k}$.

Answer 1

Angle between vectors $\overrightarrow{a},\overrightarrow{b}$ be

then $\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}$

$\cos \theta =\frac{(\hat{i}+2\hat{j}+2\hat{k})(3\hat{i}+2\hat{j}+6\hat{k})}{|\hat{i}+2\hat{j}+2\hat{k}||3\hat{i}+2\hat{j}+6\hat{k}|}$

$=\frac{3+4+12}{\sqrt{9}.\sqrt{49}}=\frac{19}{21}$

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{{19}^2}{{21}^2}}=\frac{\sqrt{{21}^2-{19}^2}}{21}=\frac{4\sqrt{5}}{21}$

Also $\sin \theta =\left|\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right|$