Question

Find the minimum distance between the lines:
$\frac{x-6}{3}=\frac{y-7}{-1}=\frac{z-4}{1}$ and $\frac{x}{-3}=\frac{y+9}{2}=\frac{z-2}{4}$

Answer 1

Equation of the given lines are
$\frac{x-6}{3}=\frac{y-7}{-1}=\frac{z-4}{1}=r...\left(1\right)$ (Let)
$\frac{x}{-3}=\frac{y+9}{2}=\frac{z-2}{4}=r^{\prime }.......\left(2\right)$ (Let)
Point on line (i) $P(3r+6,-r+7,r+4).....\left(3\right)$
Point on line (ii) $Q(-3r^{\prime },2r^{\prime }-9,4r^{\prime }+2)......\left(4\right)$
Direction ratios of the line $PQ$
$3r+6+3r^{\prime }1,-r+7-2r^{\prime }+9,r+4-4r^{\prime }-2$
If the minimum distance between lines is $PQ$ then $PQ$ will be perpendicular to both the lines.
$3(3r+3r^{\prime }+6)-1(-r-2r^{\prime }+16)+1(r-4r^{\prime }+2)=0$
$9r+9r^{\prime }+18+r+2r^{\prime }-16-r-4r^{\prime }+2=0$
$11r+7r^{\prime }+4=\mathrm{0.....}\left(5\right)$
$-3(3r+3r^{\prime }+6)+2(-r-2r^{\prime }+16)+4(r-4r^{\prime }+2)=0$
$7r+29r^{\prime }-22=\mathrm{0.......}\left(5\right)$
From equation (5) and $\left(6\right)$
$\frac{r}{-154-116}=\frac{r^{\prime }}{28+242}=\frac{1}{319-49}$
$\frac{r}{-270}=\frac{r^{\prime }}{270}=\frac{1}{270}$
$r=-1$ and $r^{\prime }=1$
Putting values in equation (3) and (4)
$P(3,8,3)$ and $Q(3,-7,6)$
Minimum distance$=PQ$
$=\sqrt{{(3+3)}^2+{(8+7)}^2+{(3-7)}^2}$
$=\sqrt{36+229+9}=3\sqrt{30}$