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Mathematics Question on Three Dimensional Geometry

Find the shortest distance between the lines x+17=y+1−6=z+11\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and x−31=y−5−2=z−71\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Answer

The given lines are x+17=y+1−6=z+11\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and x−31=y−5−2=z−71\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

It is known that the shortest distance between the two lines,
x−x1a1=y−y1b1=z−z1c1\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} and x−x2a2=y−y2b2=z−z2c2\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}, is given by,

d=∣x2−x1y2−y1z2−z1\a1b1c1\a2b2c2∣(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2\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,
∣x2−x1y2−y1z2−z1\a1b1c1\a2b2c2∣\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}

=∣468\7−61\1−21∣\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
⇒(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2\Rightarrow \sqrt{(b_1c_2-b_2c_1)^2+(c1a2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}

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

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

=116\sqrt{116}
=2292\sqrt{29}

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

d=−116229\frac{-116}{2\sqrt{29}}

=−5829\frac{-58}{\sqrt{29}}

=−2∗2929\frac{-2*29}{\sqrt{29}}

=-229\sqrt{29}

Since, the distance is always non-negative, the distance between the given lines is 229\sqrt{29} units.