Question

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

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

Answer 1

Answer: (C)

$\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}\text{ and }\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$
$\Rightarrow \text{ Shortest distance }=\frac{(\vec{b}-\vec{a})\cdot (\vec{p}\times \vec{q})}{|\vec{p}\times \vec{q}|}$
$\text{ S.D. }=(-\hat{i}+2\hat{j}-\hat{k})\cdot \frac{(\vec{p}\times \vec{q})}{|\vec{p}\times \vec{q}|}$
$\left\{\vec{p}\times \vec{q}\equiv \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& \frac{1}{2}& \frac{-1}{12}\\ 1& 1& \frac{1}{6}\end{array}\right|=\frac{1}{6}\hat{i}-\frac{1}{4}\hat{j}+\frac{1}{2}\hat{k}\text{ or }2\hat{i}-3\hat{j}+6\hat{k}\right\}$
$\text{ S.D. }=\frac{(-\hat{i}+2\hat{j}-\hat{k})\cdot (2\hat{i}-3\hat{j}+6\hat{k})}{\sqrt{2^2+3^2+6^2}}=\left|\frac{-14}{7}\right|=2$