Question

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

Answer 1

Let $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$

$\therefore \overrightarrow{a}\times \overrightarrow{b}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 3& -2& 1\end{array}\right]=4\hat{i}+8\hat{j}+4\hat{k}$

Since $sin\theta =\frac{|\overrightarrow{a}\times \overrightarrow{b}|}{|\overrightarrow{a}||\overrightarrow{b}|}\therefore sin\theta =\frac{\sqrt{16+64+16}}{\sqrt{1+4+9}\sqrt{9+4+1}}=\frac{4\sqrt{6}}{14}=\frac{2\sqrt{6}}{7}$