Find the angle between the following pairs of lines:

Solution:

i. Let $\vec{b}_{1}$ and $\vec{b}_{2}$ 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.

$\therefore \vec{b}_{1}=2 \hat{i}+5 \hat{j}-3 \hat{k}$ and $\vec{b}_{2}=-\hat{i}+8 \hat{j}+4 \hat{k}$

$\left|\vec{b}_{1}\right|=\sqrt{(2)^{2}+(5)^{2}+(-3)^{2}}=\sqrt{38}$

$\left|\vec{b}_{2}\right|=\sqrt{(-1)^{2}+(8)^{2}+(4)^{2}}=\sqrt{81}=9$

$\vec{b}_{1} \cdot \vec{b}_{2}=(2 \hat{i}+5 \hat{j}-3 \hat{k}) \cdot(-\hat{i}+8 \hat{j}+4 \hat{k})$

$=2(-1)+5 \times 8+(-3) \cdot 4$

$=-2+40-12$

$=26$

The angle, Q, between the given pair of lines is given by the relation,

$\cos Q=\left|\frac{\vec{b}_{1} \cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}\right|$

$\Rightarrow \cos Q=\frac{26}{9 \sqrt{38}}$

$\Rightarrow Q=\cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)$

(ii) Let $\vec{b}_{1}, \vec{b}_{2}$ be the vectors parallel to the given pair of lines, $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-5}{1}=\frac{z-3}{8}$, respectively.

$\vec{b}_{1}=2 \hat{i}+2 \hat{j}+\hat{k}$

$\vec{b}_{2}=4 \hat{i}+\hat{j}+8 \hat{k}$

$\therefore\left|\vec{b}_{1}\right|=\sqrt{(2)^{2}+(2)^{2}+(1)^{2}}=\sqrt{9}=3$

$\left|\vec{b}_{2}\right|=\sqrt{4^{2}+1^{2}+8^{2}}=\sqrt{81}=9$

$\vec{b}_{1} \cdot \vec{b}_{2}=(2 \hat{i}+2 \hat{j}+\hat{k}) \cdot(4 \hat{i}+\hat{j}+8 \hat{k})$

$=2 \times 4+2 \times 1+1 \times 8$

$=8+2+8$

$=18$

If $Q$ is the angle between the given pair of lines, then $\cos Q=\left|\vec{b}_{1} \cdot \vec{b}_{2}\right||| \vec{b}_{1}|| \vec{b}_{2}||$

$\Rightarrow \cos Q=\frac{18}{3 \times 9}=\frac{2}{3}$

$\Rightarrow Q=\cos ^{-1}\left(\frac{2}{3}\right)$