Question

Find the angle between the following pairs of lines:
$\overrightarrow{r}=3\hat{i}+\hat{j}-2\hat{k}+\lambda (\hat{i}+\hat{j}-2\hat{k})$ and $\overrightarrow{r}=2\hat{i}-\hat{j}-56\hat{k}+\mu (3\hat{i}-5\hat{j}-4\hat{k})$.

Answer 1

The given lines are parallel to the vectors, $\overrightarrow{b_1}=(\hat{i}-\hat{j}-2\hat{k})$ and $\overrightarrow{b_2}=(3\hat{i}-5\hat{j}-4\hat{k})$ respectively.

Therefore, $\left|\overrightarrow{b_1}\right|=\sqrt{{\left(1\right)}^2+{(-1)}^2+{(-2)}^2}=\sqrt{6}$
and $\left|\overrightarrow{b_2}\right|=\sqrt{{\left(3\right)}^2+{(-5)}^2+{(-4)}^2}=5\sqrt{2}$
Thus $\overrightarrow{b_1}.\overrightarrow{b_2}=(\hat{i}-\hat{j}-2\hat{k}).(3\hat{i}-5\hat{j}-4\hat{k})$
$=1.3-1(-5)-2(-4)$

$=3+5+8=16$
If $\theta$ is the angle between the given lines then,
$\cos \theta =\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}\right|$
$\Rightarrow$ $\cos \theta =\frac{16}{\sqrt{6}.5\sqrt{2}}=\frac{16}{\sqrt{2}.\sqrt{3}.5\sqrt{2}}=\frac{16}{10\sqrt{3}}$
$\Rightarrow$ $\cos \theta =\frac{8}{5\sqrt{3}}$
$\Rightarrow$ $\theta ={\cos }^{-1}\left(\frac{8}{5\sqrt{3}}\right)$

This the required angle between the given pair of lines.