Question

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

Answer 1

$\Rightarrow \overline{i}+2\overline{j}+2\overline{k},3\overline{i}+2\overline{j}+6\overline{k}$

$\Rightarrow \cos \theta =\frac{3\left(1\right)+2\left(2\right)+2\left(6\right)}{\sqrt{1+2^2+2^2}\sqrt{3^2+2^2+6^2}}=\frac{3+4+12}{\sqrt{9}\sqrt{49}}=\frac{19}{3\times 7}$

$\Rightarrow \cos \theta =\frac{19}{21}$

$\Rightarrow {\cos }^2\theta =\frac{{19}^2}{{21}^2}$

$\Rightarrow {\sin }^2\theta =1-\frac{{19}^2}{{21}^2}=\frac{441-361}{441}=\frac{80}{441}$ .

$\Rightarrow \sin \theta =\frac{\sqrt{10\times 5}}{21}=\frac{4\sqrt{5}}{21}$ .