Mathematics · 2020
JEE Main · 6 September 2020, Shift 1 · Q67
The shortest distance between the lines (x-1)/0=(y+1)/-1=z/1 and x+y+z+1=0, 2 x-y+z+3=0 is:
The shortest distance between the lines $\displaystyle \frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1} \quad$ and $\displaystyle x+y+z+1=0$, $\displaystyle 2 x-y+z+3=0$ is :
Official answer
(3)
$\displaystyle \frac{1}{\sqrt{3}}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.