Question

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

Answer 1

The given lines are \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)

It is known that the shortest distance between the two lines,
\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) and \(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\), is given by,

d=\(\frac{\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}\)....(1)

Comparing the given equations, we obtain,
x₁=-1, y₁=-1, z₁=-1
a₁=7, b₁=-6, c₁=1
x₂=3, y₂=5, z₂=7
a₂=1, b₂=-2, c₂=1

Then,
\(\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}\)

=\(\begin{vmatrix}4&6&8\\7&-6&1\\1&-2&1\end{vmatrix}\)

=4(-6+2)-6(7-1)+8(-14+6)
=-16-36-64
=-116
\(\Rightarrow \sqrt{(b_1c_2-b_2c_1)^2+(c1a2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}\)

=\(\sqrt{(-6+2)^2+(1+7)^2+(-14+6)^2}\)

=\(\sqrt{16+36+64}\)

=\(\sqrt{116}\)
=\(2\sqrt{29}\)

Substituting all the values in equation(1), we obtain

d=\(\frac{-116}{2\sqrt{29}}\)

=\(\frac{-58}{\sqrt{29}}\)

=\(\frac{-2*29}{\sqrt{29}}\)

=-2\(\sqrt{29}\)

Since, the distance is always non-negative, the distance between the given lines is 2\(\sqrt{29}\) units.