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Mathematics Question on Three Dimensional Geometry

The shortest distance between the lines x−1=y1=z1\frac{x}{-1}=\frac{y}{1}=\frac{z}{1} and x−30=y+31=z−3−1\frac{x-3}{0}=\frac{y+3}{1}=\frac{z-3}{-1} is

A

6\sqrt{6}

B

66

C

232\sqrt{3}

D

323\sqrt{2}

Answer

6\sqrt{6}

Explanation

Solution

Given, lines are x−1=y1=z1\frac{x}{-1}=\frac{y}{1}=\frac{z}{1} and x−30=y+31=z−3−1\frac{x-3}{0}=\frac{y+3}{1}=\frac{z-3}{-1} Here, x1=0, y1=0, z1=0{{x}_{1}}=0,\,{{y}_{1}}=0,\,{{z}_{1}}=0 l1=−1,m1,=1,n1=1{{l}_{1}}=-1,{{m}_{1}},=1,{{n}_{1}}=1 and x2=3, y2=−3, z2=3{{x}_{2}}=3,\,{{y}_{2}}=-3,\,{{z}_{2}}=3 l2=0,m2=1,n2=−1{{l}_{2}}=0,{{m}_{2}}=1,{{n}_{2}}=-1
Now, shortest distance d=∣∣x2−x1y2−y1z2−z1 I1m1n1 I2m2n2 ∣(m1n2−m2n1)2+(n1I2−n2I1)2+(I1m2−I2m1)2∣d=\left| \frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{I}_{1}} & {{m}_{1}} & {{n}_{1}} \\\ {{I}_{2}} & {{m}_{2}} & {{n}_{2}} \\\ \end{matrix} \right|}{\sqrt{{{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}+{{({{n}_{1}}{{I}_{2}}-{{n}_{2}}{{I}_{1}})}^{2}}+{{({{I}_{1}}{{m}_{2}}-{{I}_{2}}{{m}_{1}})}^{2}}}} \right|
=∣∣3−33 −111 01−1 ∣(−1−1)2+(0−1)2+(−1−0)2∣=\left| \frac{\left| \begin{matrix} 3 & -3 & 3 \\\ -1 & 1 & 1 \\\ 0 & 1 & -1 \\\ \end{matrix} \right|}{\sqrt{{{(-1-1)}^{2}}+{{(0-1)}^{2}}+{{(-1-0)}^{2}}}} \right|
=∣3(−1−1)+3(1−0)+3(−1−0)4+1+1∣=\left| \frac{3(-1-1)+3(1-0)+3(-1-0)}{\sqrt{4+1+1}} \right|
=∣−6+3−36∣=∣−6∣=6=\left| \frac{-6+3-3}{\sqrt{6}} \right|=|-\sqrt{6}|=\sqrt{6}