Verified Solution • 2025–26 Session moderate

QJEEM12Ma9513: JEE Main 2012 Mathematics - Class 12 JEEM12Ma9513 Solution: If the line and intersect, then is equal to :

If the line x−12=y+13=z−14{{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4} and x−31=y−k2=z1{{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1} intersect, then kk is equal to :

Multiple Choice Options

  • A −1-1
  • B 29{2 \over 9}
  • C 92{9 \over 2}
  • D 00

Step-by-Step Verified Solution

Given lines in vector form are

r→=(i^−j→+k→)+λ(2i→+3j→+4j→)\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)

and r→=(3i^+kj^)+μ(i^+2j^+k^)\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\widehat i + 2\widehat j + \widehat k} \right)

These will intersect if shortest distance between them =0=0

i.e. (a→2−a→1).b1→×b→2=0\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).\overrightarrow {{b_1}} \times {\overrightarrow b _2} = 0

⇒|3−1k+1−1234121|=0

⇒2(−5)−(k+1)(−2)−1(1)=0\Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0

⇒k=9/2\Rightarrow k = 9/2

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