Question

Find the shortest distance between the lines
$\frac{x+1}{6}=\frac{y+1}{-5}=\frac{z+1}{2}$ and $\frac{x-3}{2}=\frac{y-5}{-3}=\frac{z-7}{2}$

Answer 1

The lines are $\frac{x+1}{6}=\frac{y+1}{-5}=\frac{z+1}{2}$ ..........$\left(1\right)$

and $\frac{x-3}{2}=\frac{y-5}{-3}=\frac{z-7}{2}$ ..........$\left(2\right)$

Here $x_1=-1,y_1=-1,z_1=-1$ and $a_1=6,b_1=-5,c_1=2$

$x_2=3,y_2=5,z_2=7$ and $a_2=2,b_2=-3,c_2=2$

Shortest distance between the lines is

$d=\frac{\left|\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|\right|}{\sqrt{{(b_1{c}_2-b_2{c}_1)}^2+{(c_1{a}_2-c_2{a}_1)}^2+{(a_1{b}_2-a_2{b}_1)}^2}}$

$=\frac{\left|\left|\begin{array}{ccc}3+1& 5+1& 7+1\\ 6& -5& 2\\ 2& -3& 2\end{array}\right|\right|}{\sqrt{{(-10+6)}^2+{(4-12)}^2+{(-18+10)}^2}}$

$=\frac{\left|\left|\begin{array}{ccc}4& 6& 8\\ 6& -5& 2\\ 2& -3& 2\end{array}\right|\right|}{\sqrt{{(-4)}^2+{(-8)}^2+{(-8)}^2}}$

$=\left|\frac{4(-10+6)-6(12-4)+8(-18+10)}{\sqrt{16+64+64}}\right|$

$=\left|\frac{4(-4)-6\left(8\right)+8(-8)}{\sqrt{144}}\right|$

$=\left|\frac{-16-48-64}{12}\right|$

$=\left|\frac{-128}{12}\right|=\frac{32}{3}$units