Question

The shortest distance between the lines $x+\alpha =2y=12z$ and $x=y+29=6(z-\alpha )$

• A. $\alpha$
• B. $\frac{1}{2}\alpha$
• C. $\frac{12\alpha +40}{\sqrt{148}}$
• D. $3\alpha$

Answer 1

Answer: (C)

$\frac{12\alpha +40}{\sqrt{148}}$

$x+\alpha =2y=12z$ and $x=y+29=6(z-\alpha )$

$\frac{x+\alpha }{1}=\frac{y}{(\frac{1}{2})}=\frac{z}{(\frac{1}{12})}$ and $\frac{x}{1}=\frac{y+20}{1}=\frac{z-\alpha }{(\frac{1}{6})}$

Shortest distance is $\frac{\left|\begin{array}{ccc}-\alpha -0& 0+20& 0-\alpha \\ 1& \frac{1}{2}& \frac{1}{12}\\ 1& 1& \frac{1}{6}\end{array}\right|}{\sqrt{(\frac{1}{12}-\frac{1}{12}{)}^2+(1-\frac{1}{2}{)}^2+(\frac{1}{6}-\frac{1}{12}{)}^2}}=\frac{-\alpha \left(0\right)-20\times \frac{1}{12}-\alpha \times \frac{1}{2}}{\frac{\sqrt{148}}{2\times 12}}=\frac{12\alpha +40}{\sqrt{148}}$