Question

If θ is the angle between two vectors $\hat{i}-2\hat{j}+3\hat{k}\mathrm{and}3\hat{i}-2\hat{j}+\hat{k},\mathrm{find}\sin \theta$.

Answer 1

Let $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$. If θ is the angle between them. Then,
$\cos \theta =\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}$
Now,
$$\begin{gathered} \overrightarrow{a}·\overrightarrow{b}=(\hat{i}-2\hat{j}+3\hat{k})·(3\hat{i}-2\hat{j}+\hat{k}) \\ =3+4+3=10 \end{gathered}$$

$\left|\overrightarrow{a}\right|=\sqrt{1+4+9}=\sqrt{14}\mathrm{and}\left|\overrightarrow{b}\right|=\sqrt{9+4+1}=\sqrt{14}$
$$\begin{gathered} \therefore \cos \theta =\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|} \\ \Rightarrow \cos \theta =\frac{10}{\sqrt{14}\sqrt{14}}=\frac{10}{14}=\frac{5}{7} \end{gathered}$$
Now,
$$\begin{gathered} \sin \theta =\sqrt{1-{\cos }^2\theta } \\ =\sqrt{1-{\left(\frac{5}{7}\right)}^2} \\ =\sqrt{\frac{24}{49}}=\frac{2\sqrt{6}}{7} \end{gathered}$$