Question

Find the angle between the pair of lines given by
i) $\overrightarrow{r}=2\hat{i}-5\hat{j}+\hat{k}+\lambda (3\hat{i}+2\hat{j}+6\hat{k})$ and $\overrightarrow{r}=7\hat{i}-6\hat{k}+\mu (\hat{i}+2\hat{j}+2\hat{k})$
ii) $\overrightarrow{r}=3\hat{i}+\hat{j}-2\hat{k}+\lambda (\hat{i}-\hat{j}-2\hat{k})$ and $\overrightarrow{r}=2\hat{i}-\hat{j}-56\hat{k}+\mu (3\hat{i}-5\hat{j}-4\hat{k})$

Answer 1

1)Given lines:

$\overrightarrow{r}=2\hat{i}-5\hat{j}+\hat{k}+\lambda (3\hat{i}+2\hat{j}+6\hat{k})$ and $\overrightarrow{r}=7\hat{i}-6\hat{k}+\mu (\hat{i}+2\hat{j}+2\hat{k})$

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

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

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

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

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

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

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

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

Hence, the angle between two lines is ${\cos }^{-1}\left(\frac{19}{21}\right)$

ii)Given lines:

$\overrightarrow{r}=3\hat{i}+\hat{j}-2\hat{k}+\lambda (\hat{i}-\hat{j}-2\hat{k})$ and $\overrightarrow{r}=2\hat{i}-\hat{j}-56\hat{k}+\mu (3\hat{i}-5\hat{j}-4\hat{k})$

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

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

Here, ${\overrightarrow{b}}_1=\hat{i}-\hat{j}-2\hat{k}$ and ${\overrightarrow{b}}_2=3\hat{i}-5\hat{j}-4\hat{k}$ So,

$\cos \theta =\left|\frac{(\hat{i}-\hat{j}-2\hat{k}).(3\hat{i}-5\hat{j}-4\hat{k})}{|\hat{i}-\hat{j}-2\hat{k}|.|3\hat{i}-5\hat{j}-4\hat{k}|}\right|$

$\Rightarrow \cos \theta =\left|\frac{3+5+8}{\sqrt{1^2+1^2+2^2}\sqrt{3^2+5^2+4^2}}\right|$

$\Rightarrow \cos \theta =\left|\frac{16}{\sqrt{6}\sqrt{50}}\right|$

$\Rightarrow \cos \theta =\frac{16}{10\sqrt{3}}=\frac{8}{5\sqrt{3}}$

$\therefore \theta ={\cos }^{-1}\left(\frac{8}{5\sqrt{3}}\right)$
Hence, the angle between two lines is ${\cos }^{-1}\left(\frac{8}{5\sqrt{3}}\right)$