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

The shortest distance between the lines $1+x=2 y=-12 z$ and $x=y+2=6 z-6$ is
  1. $1$ unit
  2. $4$ units
  3. $2$ units
  4. $3$ units

Solution

Shortest distance between the lines $\begin{array}{c} \frac{x+1}{1}=\frac{y}{\left(\frac{1}{2}\right)}=\frac{z}{\left(\frac{-1}{12}\right)} \text { and } \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\left(\frac{1}{6}\right)} \text { is } \\ \begin{array}{c} 0+1-2-0 \\ 1 \\ \frac{1}{2} & \frac{-1}{12} \\ 1 & \frac{1}{6} \end{array} \mid \\ d=\frac{\sqrt{\left(\frac{1}{6}+\frac{1}{12}\right)^{2}+\left(1-\frac{1}{2}\right)^{2}+\left(\frac{1}{12}+\frac{1}{12}\right)^{2}}}{12} \end{array}$ $=\frac{\left|\begin{array}{ccc}1 & -2 & 1 \\ 1 & \frac{1}{2} & \frac{-1}{12} \\ 1 & 1 & \frac{1}{6}\end{array}\right|}{\sqrt{\left(\frac{1}{4}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{6}\right)^{2}}}=\frac{(1)\left(\frac{1}{12}+\frac{1}{12}\right)+2\left(\frac{1}{6}+\frac{1}{12}\right)+\left(1-\frac{1}{2}\right) \mid}{\sqrt{\frac{1}{16}+\frac{1}{4}+\frac{1}{36}}}$ $=\frac{\left|\frac{1}{6}+\frac{1}{2}+\frac{1}{2}\right|}{\sqrt{\frac{9+36+4}{144}}}=\frac{\left|\frac{7}{6}\right|}{\sqrt{\frac{49}{144}}}=\frac{7}{6} \times \frac{12}{7}=2$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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