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

Answer

$\vec{\text{a}}\cdot\vec{\text{b}}=|\vec{\text{a}}||\vec{\text{b}}|\cos\theta$
$\vec{\text{a}}\cdot\vec{\text{b}}=(\hat{\text{i}}-2\hat{\text{j}}+3\vec{\text{k}})\cdot(3\hat{\text{i}}-2\hat{\text{i}}+\hat{\text{k}})$
$=3+4+3$
$=10$
$10=\big(\sqrt{1+4+9}\big)\big(\sqrt{9+4+1}\big)\cos\theta$
$\cos\theta=\frac{10}{\sqrt{14}\sqrt{14}}$
$\cos\theta=\frac{10}{14}$
$\sin\theta=\sqrt{1-\cos^2\theta}$
$=\sqrt{{1-\frac{25}{49}}}=\sqrt{\frac{49-25}{49}}$
$=\frac{\sqrt{24}}{7}=\frac{2\sqrt{6}}{7}$

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