Question

Find distance between the lines $\frac{x-4}{1}=\frac{y-5}{2}=\frac{z}{2}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-2}{4}$

Answer 1

Line1:$\frac{x-4}{1}=\frac{y-5}{2}=\frac{z}{2}$ , Line2:$\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-2}{4}$

${\overrightarrow{a}}_1=4\hat{i}+5\hat{j}+0\hat{k};{\overrightarrow{a}}_2=-1\hat{i}-3\hat{j}+2\hat{k}{\overrightarrow{b}}_1=1\hat{i}+2\hat{j}+2\hat{k};{\overrightarrow{b}}_2=2\hat{i}+4\hat{j}+4\hat{k}$

both the lines are parallel $\Rightarrow$ distance between lines$=$ perpendicular distance from ${\overrightarrow{a}}_1$ on line2.

$\therefore (2r-1,4r-3,4r+2)$ be the point of foot of $\perp$

$\therefore (2r-1-4,4r-3-5,4r+2-0)$ should be $\perp$ to ${\overrightarrow{b}}_2$

$\therefore (2r-5)2+(4r-8)\times 4+(4r+2)\times 4=0\Rightarrow r=\frac{17}{18}$

$\therefore$ distance $=\sqrt{{(2r-5)}^2+{(4r-8)}^2+{(4r+2)}^2}=\frac{2}{3}\sqrt{137}$