Verified Solution • 2025–26 Session moderate

QJEEM22Ma61d3: JEE Main 2022 Mathematics - Class 12 JEEM22Ma61d3 Solution: The shortest distance between the lines and , is :

The shortest distance between the lines

x−32=y−23=z−1−1{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}} and x+32=y−61=z−53{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3} , is :

Multiple Choice Options

  • A 185{{18} \over {\sqrt 5 }}
  • B 2235{{22} \over {3\sqrt 5 }}
  • C 4635{{46} \over {3\sqrt 5 }}
  • D 636\sqrt 3

Step-by-Step Verified Solution

L1:x−32=y−23=z−1−1{L_1}:{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}

L2:x+32=y−61=z−53{L_2}:{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}

Now, p→×q→=|i^j^k^23−1213|=10i^−8j^−4k^

and a→2−a→1=6i^−4j^−4k^{\overrightarrow a _2} - {\overrightarrow a _1} = 6\widehat i - 4\widehat j - 4\widehat k

∴\therefore S.D.=∣60+32+16100+64+16∣=108180=185S.D. = \left| {{{60 + 32 + 16} \over {\sqrt {100 + 64 + 16} }}} \right| = {{108} \over {\sqrt {180} }} = {{18} \over {\sqrt 5 }}

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