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Mathematics Question on Shortest Distance between Two Lines

The shortest distance between the lines x−32=y−23=z−1−1\frac{x−3}{2}=\frac{y−2}{3}=\frac{z−1}{−1} and x+32=y−61=z−53\frac{x+3}{2}=\frac{y-6}{1}=\frac{z−5}{3} is

A

185\frac{18}{\sqrt5}

B

2235\frac{22}{3\sqrt5}

C

4635\frac{46}{3\sqrt5}

D

636\sqrt3

Answer

185\frac{18}{\sqrt5}

Explanation

Solution

L1L_1 : x−32=y−23=z−1−1\frac{x−3}{2}=\frac{y−2}{3}=\frac{z−1}{−1}

L2L_2 : x+32=y−61=z−53\frac{x+3}{2}=\frac{y-6}{1}=\frac{z−5}{3}

Now,

p→×q→\overrightarrow p×\overrightarrow q = ∣i^j^k^ 23−1 213∣=10i^−8j^−4k^\begin{vmatrix}\hat i& \hat j& \hat k\\\ 2 &3 &−1 \\\ 2 & 1& 3 \end{vmatrix} =10\hat i−8\hat j−4\hat k

a→2−a→1=6i^−4j^−4k^\overrightarrow a_2−\overrightarrow a_1=6\hat i−4\hat j−4\hat k

∴S.D=∣60+32+16100+64+16∣∴S.D=\begin{vmatrix}\frac{60+32+16}{\sqrt{100+64+16}}\end{vmatrix}

= 108180=185\frac{108}{\sqrt{180}}=\frac{18}{\sqrt5}