If $\left|\overrightarrow{a}\right| = 3$, $\;$ $\left|\overrightarrow{b}\right| = 4$ $\;$ and $\;$ $\overrightarrow{a} \cdot \overrightarrow{b} = 9$, then find $\left|\overrightarrow{a} \times \overrightarrow{b}\right|$
Given: $\;$ $\left|\overrightarrow{a}\right| = 3$, $\;$ $\left|\overrightarrow{b}\right| = 4$, $\;$ $\overrightarrow{a} \cdot \overrightarrow{b} = 9$
By definition, $\;$ $\overrightarrow{a} \cdot \overrightarrow{b} = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \cos \theta$
i.e. $\;$ $9 = 3 \times 4 \times \cos \theta$ $\implies$ $\cos \theta = \dfrac{3}{4}$
$\therefore$ $\;$ $\sin \theta = \sqrt{1 - \cos ^2 \theta} = \sqrt{1 - \dfrac{9}{16}} = \dfrac{\sqrt{7}}{4}$
$\begin{aligned}
\text{Now, } \; \left|\overrightarrow{a} \times \overrightarrow{b}\right| & = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \sin \theta \\\\
& = 3 \times 4 \times \dfrac{\sqrt{7}}{4} = 3 \sqrt{7}
\end{aligned}$