Question

Write the vector equations of the following line and henece find the shortest distance between them:$\frac{x+1}{2}=\frac{y+1}{-6}=\frac{z+1}{1}and\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

Answer 1

$\frac{x+1}{2}=\frac{y+1}{-6}=\frac{2+1}{1}{K}_1$ ---(1)

and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{3-7}{1}K$ ----(2)

Point on line (1) is $A(2K-1,6K-1,K-1)$

Point on line (2) is $B(t+3,-2t+5,t+7)$

If AB is the shortest distance than direction ratios of AB is

$=\left(\right(2K-1)-(t+3),(-6K-1)-(-2t+5),(k-1)-(t+7\left)\right)$

i.e $(2K-t-4,-6K+2t-6,k-t-8)$

AB is $\perp$ to lines (1) & (2)

$2\ast (2K-T-4)-6(-6k+2t-6)+1(K-t-8)=0$

$4K-2t-8+36K-12t+36+K-t-8=0$

$41K-15t+20=0$ ---(3)

& $1(2K-t-4)-2(-6K+2t-6)+1(K-t-8)=0$

$2K-t-4+12K-4t+12+K-t-8=0$

$15K-6t=0$

$K=\frac{6t}{15}=\frac{2t}{5}$

$\frac{41\times 2t}{5}-15t+20=0$

$82t-75t+100=0$

$7t=-100$

$t=\frac{-1000}{7}$

$K=\frac{-40}{7}$

Point $A=(\frac{-87}{7},\frac{233}{7},\frac{-47}{7}),B(\frac{-79}{7},\frac{235}{7},\frac{-51}{7})$

Shortest distance $AB=\sqrt{(\frac{-79}{7}+\frac{87}{7}){(\frac{235}{7}-\frac{233}{7})}^2+{(\frac{-57}{7}+\frac{47}{7})}^2}$

$=\sqrt{{\left(\frac{8}{7}\right)}^2+{\left(\frac{2}{7}\right)}^2+{\left(\frac{-4}{7}\right)}^2}$

$\Rightarrow \sqrt{\frac{64+4+16}{7^2}}$

$\Rightarrow \sqrt{\frac{84}{7^2}}\Rightarrow \frac{2\sqrt{21}}{7}$

direction ratio of shortest distance line are $(\frac{-8}{7},\frac{-2}{7},\frac{4}{7})$

equation of shortest distance

$\frac{x+87/7}{-8/7}=\frac{y-233/7}{-2/7}=\frac{3+47/7}{4/7}$

or $\frac{7x+87}{-8}=\frac{7y-233}{-2}=\frac{73+47}{4}$