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

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

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

and $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.

$|\overrightarrow{a}|=\sqrt{1^2+(-2{)}^2+(3{)}^2}$

$=\sqrt{1+4+9}=\sqrt{14}$

$|\overrightarrow{b}|=\sqrt{3^2+(-2{)}^2+1^2}$

$=\sqrt{9+4+1}=\sqrt{14}$

Now, $\left|\begin{array}{c}\overrightarrow{a}\end{array}\right|\left|\begin{array}{c}\overrightarrow{b}\end{array}\right|sin\theta =\left|\begin{array}{c}\overrightarrow{a}\times \overrightarrow{b}\end{array}\right|$

$sin\theta =\frac{\left|\begin{array}{c}\overrightarrow{a}\times \overrightarrow{b}\end{array}\right|}{\left|\begin{array}{c}\overrightarrow{a}\end{array}\right|\left|\begin{array}{c}\overrightarrow{b}\end{array}\right|}=\frac{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 3& -2& 1\end{array}\right|}{\sqrt{14}\sqrt{14}}$

= $\frac{\left|\begin{array}{c}4\hat{i}+8\hat{j}+4\hat{k}\end{array}\right|}{14}$

= $\frac{\sqrt{16+64+16}}{14}=\frac{\sqrt{96}}{14}=\frac{4\sqrt{6}}{14}=\frac{2}{7}\sqrt{6}$