Question

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

• A. $\frac{3}{2}$
• B. 3
• C. 2
• D. $\frac{5}{2}$

Answer 1

Answer: (C)

$ \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-1}{\frac{1}{6}} $
$ \Rightarrow \text { Shortest distance }=\frac{(\vec{ b }-\vec{ a }) \cdot(\vec{ p } \times \vec{ q })}{|\vec{ p } \times \vec{ q }|}$
$ \text { S.D. }=(-\hat{ i }+2 \hat{ j }-\hat{ k }) \cdot \frac{(\vec{ p } \times \vec{ q })}{|\vec{ p } \times \vec{ q }|}$
$ \left\{\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 }\right\} $
$\text { S.D. }=\frac{(-\hat{ i }+2 \hat{ j }-\hat{ k }) \cdot(2 \hat{ i }-3 \hat{ j }+6 \hat{ k })}{\sqrt{2^2+3^2+6^2}}=\left|\frac{-14}{7}\right|=2$