Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

The shortest distance, between two lines
$\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$
and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$
is $d=\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|}{\sqrt{\begin{array}{rl}& {(m_1{n}_2-m_2{n}_1)}^2+{(n_1{l}_2-l_1{n}_2)}^2\\ & +{(l_1{m}_2-m_1{l}_2)}^2\end{array}}}$
Given lines are $\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}$
and $\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$
$\therefore$ $d=\frac{\left|\begin{array}{ccc}0+1& -2-0& 1-0\\ 12& 6& -1\\ 6& 6& 1\end{array}\right|}{\sqrt{{(6+6)}^2+{(-6-12)}^2+{(72-36)}^2}}$
$=\frac{\left|\begin{array}{ccc}1& -2& 1\\ 12& 6& -1\\ 6& 6& 1\end{array}\right|}{\sqrt{144+324+1296}}$
$=\frac{1(6+6)+2(12+6)+1(72-36)}{\sqrt{1764}}$
$=\frac{84}{\sqrt{1764}}=\frac{84}{42}=2$