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JEE Main 2019 · Mathematics · Three Dimensional Geometry · hard

Find the least distance between the two skew lines (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-2)/(3)=(y-4)/(4)=(z-7)/(5) is

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Find the least distance between the two skew lines x−12=y−23=z−34\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4} and x−23=y−44=z−75\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-7}{5} is
  1. A. 13\frac{1}{\sqrt3}
  2. B. 16\frac{1}{\sqrt6}
  3. C. 16\frac{1}{6}
  4. D. 12\frac{1}{\sqrt2}
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Answer: (B) 16\frac{1}{\sqrt6}

Directions d⃗1=(2,3,4), d⃗2=(3,4,5)\vec d_1=(2,3,4),\ \vec d_2=(3,4,5) give n⃗=d⃗1×d⃗2=(−1,2,−1)\vec n=\vec d_1\times\vec d_2=(-1,2,-1), ∣n⃗∣=6|\vec n|=\sqrt6. Joining vector of the points is (1,2,4)(1,2,4) and n⃗⋅(1,2,4)=−1+4−4=−1\vec n\cdot(1,2,4)=-1+4-4=-1. Distance =16=\dfrac{1}{\sqrt6}.
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