Vector Algebra

If $\overrightarrow{a} =2$, $\overrightarrow{b} = 7$ and $\overrightarrow{a} \times \overrightarrow{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$, find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.


Given: $\left|\overrightarrow{a}\right| = 2$; $\;$ $\left|\overrightarrow{b}\right| = 7$; $\;$ $\overrightarrow{a} \times \overrightarrow{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$

By definition, $\;$ $\left|\overrightarrow{a} \times \overrightarrow{b}\right| = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right| \sin \theta$

$\left|\overrightarrow{a} \times \overrightarrow{b}\right| = \sqrt{\left(3\right)^2 + \left(-2\right)^2 + \left(6\right)^2} = \sqrt{49} = 7$

$\therefore$ $\;$ $\sin \theta = \dfrac{\left|\overrightarrow{a} \times \overrightarrow{b}\right|}{\left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right|} = \dfrac{7}{2 \times 7} = \dfrac{1}{2}$

$\implies$ $\theta = \sin^{-1} \left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}$