Question

Find the shortest distance (in cm) between the lines
$\frac{x-8}{3}=\frac{y+9}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{5-z}{5}$

Answer 1

Given lines are $\frac{x-8}{3}=\frac{y+9}{-16}=\frac{z-10}{}$ ......... (i)
and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$ ....... (ii)
Any point on (i) is M(8 + 3t, -9 - 16t, 10 + 7t)
and Any point on (ii) is N(15 + 3s, 29 + 8s, 5 - 5s)
$\therefore$ Direction numbers of the line MN are
$(15+3s-8-3t,29+8s+9+16t,5-5s-10-7t)$
$\Rightarrow (7+3s-3t,38+8s+16t,-5-5s-7t)$
Now, MN will be the shortest distance between (i) and (ii)
if MN is perpendicular to both.
i.e., if $3(7+3s-3t)+(-16)(38+8s+16t)+7(-5-5s-7t)=0$
and $3(7+3s-3t)+8(38+8s+16t)+(-5)(-5-5s-7t)=0$
i.e., if
$-154s-314t-622=0$
i.e., $77s+157t+311=0$ ......... (iii)
and $98s+154t+350=0$
i.e., $49s+77t+175=0$ ......... (iv)
On solving (iii) and (iv) simultaneously, we get
$t=-1$ and $s=-2$
$t=-1$ gives $M(5,7,3)$ and
$s=-2$ gives $N(9,13,15)$
$\therefore$ The shortest distance between the given lines
$MN=\sqrt{(9-5{)}^2+(13-7{)}^2+(15-3{)}^2}$
$=\sqrt{196}=14units$