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Mathematics Question on Shortest Distance between Two Lines

The shortest distance between the lines x+1=2y=−12zx+1=2 y=-12 z and x=y+2=6z−6x=y+2=6 z-6 is

A

32\frac{3}{2}

B

3

C

2

D

52\frac{5}{2}

Answer

2

Explanation

Solution

The correct answer is (C) : 2
x+11=y12=z−112 and x1=y+21=z−616\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}} \text{ and } \frac{x}{1}=\frac{y+2}{1}=\frac{z-6}{\frac{1}{6}}
⇒Shortest distance=(b⃗−a⃗)⋅(p⃗×q⃗)∣p⃗×q⃗∣⇒ \text{Shortest distance}= \frac{(\vec{b}−\vec{a})⋅(\vec{p}\times\vec{q})}{|\vec{p}\times\vec{q}|}
⇒S.D.=(−i^+2j^−k^)⋅(p⃗×q⃗)∣p⃗×q⃗∣⇒ \text{S.D.}= (-\hat{i}+2\hat{j}-\hat{k})⋅\frac{(\vec{p}\times\vec{q})}{|\vec{p}\times\vec{q}|}
\Biggl\\{\vec{p}\times\vec{q} \equiv \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\1&\frac{1}{2}&\frac{-1}{12}\\\1&1&\frac{1}{6}\end{vmatrix}=\frac{1}{6}\hat{i}-\frac{1}{4}\hat{j}+\frac{1}{2}\hat{k} \text{ or } 2\hat{i}-3\hat{j}+6\hat{k} \Biggl\\}
⇒S.D.=(−i^+2j^−k^).(2i^−3j^+6k^)22+32+62⇒ \text{S.D.}= \frac{(-\hat{i}+2\hat{j}-\hat{k}).(\hat{2i}-3\hat{j}+6\hat{k})}{\sqrt{2^2+3^2+6^2}}