Vector Algebra

Find the angle between the lines $\;$ $\overrightarrow{r} = 5 \hat{i} - 7 \hat{j} + \mu \left(- \hat{i} + 4 \hat{j} + 2 \hat{k}\right)$ $\;$ and $\;$ $\overrightarrow{r} = -2 \hat{i} + \hat{k} + \lambda \left(3 \hat{i} + 4 \hat{k}\right)$


Let the given lines be in the direction of $\overrightarrow{u}$ and $\overrightarrow{v}$

Then, $\;$ $\overrightarrow{u} = - \hat{i} + 4 \hat{j} + 2 \hat{k}$ $\;$ and $\;$ $\overrightarrow{v} = 3 \hat{i} + 4 \hat{k}$

Let $\theta$ be the angle between the two lines.

[Note: Angle between two lines is defined as the angle between their directions.]

Now, $\;$ $\cos \theta = \left(\dfrac{\overrightarrow{u} \cdot \overrightarrow{v}}{\left|\overrightarrow{u}\right| \left|\overrightarrow{v}\right|}\right)$

$\begin{aligned} \overrightarrow{u} \cdot \overrightarrow{v} & = \left(- \hat{i} + 4 \hat{j} + 2 \hat{k}\right) \cdot \left(3 \hat{i} + 4 \hat{k}\right) \\\\ & = -3 + 8 = 5 \end{aligned}$

$\left|\overrightarrow{u}\right| = \sqrt{\left(-1\right)^2 + \left(4\right)^2 + \left(2\right)^2} = \sqrt{21}$

$\left|\overrightarrow{v}\right| = \sqrt{\left(3\right)^2 + \left(4\right)^2} = \sqrt{25} = 5$

$\therefore$ $\;$ $\cos \theta = \left(\dfrac{5}{5 \sqrt{21}}\right)$

$\therefore$ $\;$ $\theta = \cos^{-1} \left(\dfrac{1}{\sqrt{21}}\right)$