Question

Find the shortest distance between the lines
$\frac{x-2}{6}=\frac{y-2}{-5}=\frac{z-2}{2}$ and $\frac{x+4}{2}=\frac{y+3}{3}=\frac{z+5}{2}$.

Answer 1

The lines are,

$\frac{x-2}{6}=\frac{y-2}{-5}=\frac{z-2}{2}$ ------ ( 1 )

$\frac{x+4}{2}=\frac{y+3}{3}=\frac{z+5}{2}$. ------ ( 2 )

Comparing the equations with,

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_1}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$

We get,

$\Rightarrow$ $x_1=2,y_1=2,z_1=2$ and $a_1=6,b_1=-5,c_1=2$

$\Rightarrow$ $x_2=-4,y_2=-3,z_2=-5$ and $a_2=2,b_2=3,c_2=2$

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$ $=\left|\begin{array}{ccc}-6& -5& -7\\ 6& -5& 2\\ 2& 3& 2\end{array}\right|$

$=-6(-10-6)+5(12-4)-7(18+10)$

$=-6(-16)+5\left(8\right)-7\left(28\right)$

$=96+40-196$

$=-60$

$\Rightarrow$ $\sqrt{(a_1{b}_2-a_2{b}_1{)}^2+(b_1{c}_2-b_2{c}_1{)}^2+(c_1{a}_2-c_2{a}_1{)}^2}$

$\Rightarrow$ $\sqrt{\left[\right(6\left)\right(3)-(2\left)\right(-5){]}^2+[(-5)\left(3\right)-\left(3\right)\left(2\right){]}^2+\left[\right(2\left)\right(2)-(2\left)\right(6){]}^2}$

$=\sqrt{(18+10{)}^2+(-15-6{)}^2+(4-12{)}^2}$

$=\sqrt{(28{)}^2+(-21{)}^2+(-8{)}^2}$

$=\sqrt{784+441+64}$

$=\sqrt{1289}$

Shortest distance between line is $d$.

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

$\Rightarrow$ $d=\left|\frac{-60}{\sqrt{1289}}\right|$

$\Rightarrow$ $d=\frac{60}{\sqrt{1289}}$