Question

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

• A. $\frac{1}{2}$
• B. $2$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

The lines are $\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$
and $\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}$
Here, ${\vec{a}}_1=-2\hat{j}+\hat{k}$,
${\vec{b}}_1=6\hat{i}+6\hat{j}+\hat{k}$,
${\vec{a}}_2=-\hat{i}$,
${\vec{b}}_2=12\hat{i}+6\hat{j}-\hat{k}$
${\vec{b}}_1\times {\vec{b}}_2=-12\hat{i}+18\hat{j}-36\hat{k}$
Shortest distance
$=\frac{|\left({\vec{a}}_2-{\vec{a}}_1\right)\cdot \left({\vec{b}}_1\times {\vec{b}}_2\right)|}{|{\vec{b}}_1\times {\vec{b}}_2|}=\frac{|\left(-\hat{i}+2\hat{j}-\hat{k}\right)\cdot \left(-12\hat{i}+18\hat{j}-36\hat{k}\right)|}{\sqrt{{\left(-12\right)}^2+{\left(18\right)}^2+{\left(-36\right)}^2}}$
$=\frac{|12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2$