Question

Find the shortest distance between the lines:
$\frac{x+1}{4}=\frac{y-3}{-6}=\frac{z+1}{1}$ and $\frac{x+3}{3}=\frac{y-5}{2}=\frac{z-7}{6}$

Answer 1

Given equation of lines-

$L_1:\frac{x+1}{4}=\frac{y-3}{-6}=\frac{z+1}{1}$

Here,

$x_1=-1,y_1=3,z_1=-1$

$a_1=4,b_1=-6,c_1=1$

$L_2:\frac{x+3}{3}=\frac{y-5}{2}=\frac{z-7}{6}$

Here,

$x_2=-3,y_2=5,z_2=7$

$a_2=3,b_2=2,c_2=6$

As we know that shortest distance between two linesusing cartesian equation is given by-

$d=\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{(b_1{c}_2-c_1{b}_2)}^2+{(c_1{a}_2-c_2{a}_1)}^2+{(a_1{b}_2-a_2{b}_1)}^2}}.....\left(1\right)$

Now,

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

$=\left|\begin{array}{ccc}-3-(-1)& 5-3& 7-(-1)\\ 4& -6& 1\\ 3& 2& 6\end{array}\right|$

$=\left|\begin{array}{ccc}-2& 2& 8\\ 4& -6& 1\\ 3& 2& 6\end{array}\right|$

$=-2(-36-2)-2(24-3)+8(8-(-18))$

$=-76-42+208=90$

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

$=\sqrt{{(-36-2)}^2+{(3-24)}^2+{(8-(-18))}^2}$

$=\sqrt{1444+441+676}$

$=\sqrt{2561}$

Therefore,

$d=\frac{90}{\sqrt{2561}}$

Hence the shortest distance between the given lines is $\frac{90}{\sqrt{2561}}$.