Question

Find the angle between the pair of lines given by
$\overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})$
and
$\overrightarrow{r}=5\hat{i}-2\hat{j}+\mu (3\hat{i}+2\hat{j}+6\hat{k})$.

Answer 1

Given lines:
$\overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})$ and

$\overrightarrow{r}=5\hat{i}-2\hat{j}+\mu (3\hat{i}+2\hat{j}+6\hat{k})$

We know, the angle between lines

$\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and $\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ is:

$\cos \theta =\left|\frac{{\overrightarrow{b}}_1.{\overrightarrow{b}}_2}{\left|{\overrightarrow{b}}_1\right|\left|{\overrightarrow{b}}_2\right|}\right|$

Here, ${\overrightarrow{b}}_1=\hat{i}+2\hat{j}+2\hat{k}$ and ${\overrightarrow{b}}_2=3\hat{i}+2\hat{j}+6\hat{k},$

$\therefore \cos \theta =\left|\frac{(\hat{i}+2\hat{j}+2\hat{k}).(3\hat{i}+2\hat{j}+6\hat{k})}{|\hat{i}+2\hat{j}+2\hat{k}||3\hat{i}+2\hat{j}+6\hat{k}|}\right|$

$\Rightarrow \cos \theta =\left|\frac{3+4+12}{\sqrt{1^2+2^2+2^2}\sqrt{3^2+2^2+6^2}}\right|$

$\Rightarrow \cos \theta =\left|\frac{19}{\sqrt{9}\sqrt{49}}\right|$

$\Rightarrow \cos \theta =\frac{19}{3\times 7}=\frac{19}{21}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{19}{21}\right)$

Hence,angle between given pair of lines is

${\cos }^{-1}\left(\frac{19}{21}\right)$