Question

Find the angle between the following pair of lines:
(i) $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$

(ii) $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$

Answer 1

(i)

Let $\overrightarrow{b_1}$ and $\overrightarrow{b_2}$ be the vectors parallel to the pair of lines,
$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and
$\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$ respectively,
then $\overrightarrow{b_1}=(\widehat{2i}+5\hat{j}-3\hat{k})$ and $\overrightarrow{b_2}=(-\hat{i}+8\hat{j}+4\hat{k})$
$\left|\overrightarrow{b_1}\right|=\sqrt{{\left(2\right)}^2+{\left(5\right)}^2+{(-3)}^2}=\sqrt{38}$
$\left|\overrightarrow{b_2}\right|=\sqrt{{(-1)}^2+{\left(8\right)}^2+{\left(4\right)}^2}=\sqrt{81}=9$
$\overrightarrow{b_1}.\overrightarrow{b_2}=(\widehat{2i}+5\hat{j}-3\hat{k}).(-\hat{i}+8\hat{j}+4\hat{k})$
$=2(-1)+5\times 8+(-3).4=-2+40-12=26$
The angle $\theta$ between the given pair of lines is given by the relation,
$\Rightarrow \cos \theta =\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}\right|$
$\Rightarrow$ $\cos \theta =\frac{26}{9\sqrt{38}}$
$\Rightarrow$ $\theta ={\cos }^{-1}\left(\frac{26}{9\sqrt{38}}\right)$

(ii)

We have,
$\overrightarrow{b_1}=(\widehat{2i}+2\hat{j}+\hat{k})$
$\overrightarrow{b_2}=(4\hat{i}+\hat{j}+8\hat{k})$
$\therefore$ $\left|\overrightarrow{b_1}\right|=\sqrt{{\left(2\right)}^2+{\left(2\right)}^2+{\left(1\right)}^2}=\sqrt{9}=3$
$\left|\overrightarrow{b_2}\right|=\sqrt{{\left(4\right)}^2+{\left(1\right)}^2+{\left(8\right)}^2}=\sqrt{81}=9$
$\overrightarrow{b_1}.\overrightarrow{b_2}=(\widehat{2i}+2\hat{j}+\hat{k}).(4\hat{i}+\hat{j}+8\hat{k})$
$=2\times 4+2\times 1+1\times 8=8+2+8=18$
If $\theta$ is the angle between the given pair of lines, then $\cos \theta =\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}\right|$
$\Rightarrow$ $\cos \theta =\frac{18}{3\times 9}=\frac{2}{3}$
$\Rightarrow$ $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$