Find the value of $m$ for which the vectors $\overrightarrow{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k}$ and $\overrightarrow{b} = \hat{i} + m \hat{j} + 3 \hat{k}$ are
- perpendicular
- parallel
If two vectors are perpendicular (angle between the vectors is $\theta = 90^{\circ}$), then $\overrightarrow{P} \cdot \overrightarrow{Q} = \left|\overrightarrow{P}\right| \left|\overrightarrow{Q}\right| \cos \left(90^{\circ}\right) = 0$
If two vectors are parallel (angle between the vectors is $\theta = 0^{\circ}$), then $\overrightarrow{P} \cdot \overrightarrow{Q} = \left|\overrightarrow{P}\right| \left|\overrightarrow{Q}\right| \cos \left(0^{\circ}\right) = \left|\overrightarrow{P}\right| \left|\overrightarrow{Q}\right|$
Now,
$\begin{aligned}
\overrightarrow{a} \cdot \overrightarrow{b} & = \left(3 \hat{i} + 2 \hat{j} + 9 \hat{k}\right) \cdot \left(\hat{i} + m \hat{j} + 3 \hat{k}\right) \\\\
& = 3 + 2m + 27 \\\\
& = 30 + 2m
\end{aligned}$
and
$\left|\overrightarrow{a}\right| = \sqrt{\left(3\right)^2 + \left(2\right)^2 + \left(9\right)^2} = \sqrt{94}$
$\left|\overrightarrow{b}\right| = \sqrt{\left(1\right)^2 + \left(m\right)^2 + \left(3\right)^2} = \sqrt{10 + m^2}$
- When $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular, $\overrightarrow{a} \cdot \overrightarrow{b} = 0$
i.e. $\;$ $30 + 2m = 0$ $\implies$ $m = - 15$ - When $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel, $\overrightarrow{a} \cdot \overrightarrow{b} = \left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right|$
i.e. $\;$ $30 + 2m = \sqrt{94} \times \sqrt{10 + m^2}$
i.e. $\;$ $900 + 4 m^2 + 120 m = 940 + 94 m^2$
i.e. $\;$ $90 m^2 - 120 m + 40 = 0$
i.e. $\;$ $9m^2 - 12 m + 4 = 0$
i.e. $\;$ $\left(3m - 2\right)^2 = 0$
i.e. $\;$ $3m - 2 = 0$ $\implies$ $m = \dfrac{2}{3}$