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Question: Answered & Verified by Expert
The shortest distance between the lines $x=y+2=6 z-6$ and $x+1=2 y=-12 z$ is
MathematicsThree Dimensional GeometryVITEEEVITEEE 2021
Options:
  • A $\frac{1}{2}$
  • B 2
  • C 1
  • D $\frac{3}{2}$
Solution:
1696 Upvotes Verified Answer
The correct answer is: 2
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, $\vec{a}_1=-2 \hat{j}+\hat{k}, \vec{b}_1=6 \hat{i}+6 \hat{j}+\hat{k}, \vec{a}_2=-\hat{i}$,
$$
\begin{aligned}
& \vec{b}_2=12 \hat{i}+6 \hat{j}-\hat{k} \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & k \\
6 & 6 & 1 \\
12 & 6 & -1
\end{array}\right|=-12 \hat{i}+18 \hat{j}-36 \hat{k}
\end{aligned}
$$
Shortest distance $=\frac{\left|\left(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1\right) \cdot\left(\overrightarrow{\mathrm{b}}_1-\overrightarrow{\mathrm{b}}_2\right)\right|}{\left|\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right|}$

$$
\begin{aligned}
& =\frac{|(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(-12 i+18 \hat{j}-36 \hat{k})|}{\sqrt{(-12)^2+(18)^2+(-36)^2}} \\
& =\frac{|+12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2
\end{aligned}
$$

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