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JEE Advanced 2020 Paper 1 · Mathematics · Three Dimensional Geometry · medium

Let L1:(x-1)/(1)=(y-2)/(-1)=(z-3)/(2) and L2:(x-2)/(2)=(y-1)/(1)=(z)/(-1). Let Pi be the plane containing L1 and parallel to L2. The perpendicular…

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Let L1:x−11=y−2−1=z−32L_1:\dfrac{x-1}{1}=\dfrac{y-2}{-1}=\dfrac{z-3}{2} and L2:x−22=y−11=z−1L_2:\dfrac{x-2}{2}=\dfrac{y-1}{1}=\dfrac{z}{-1}. Let Π\Pi be the plane containing L1L_1 and parallel to L2L_2. The perpendicular distance of the origin from Π\Pi is
  1. A. 1535\dfrac{15}{\sqrt{35}}
  2. B. 335\dfrac{3}{\sqrt{35}}
  3. C. 1835\dfrac{18}{\sqrt{35}}
  4. D. 2135\dfrac{21}{\sqrt{35}}
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Answer: (C) 1835\dfrac{18}{\sqrt{35}}

The plane Π\Pi has normal n⃗=d⃗1×d⃗2\vec n=\vec d_1\times\vec d_2 where d⃗1=(1,−1,2)\vec d_1=(1,-1,2) and d⃗2=(2,1,−1)\vec d_2=(2,1,-1). Computing, n⃗=((−1)(−1)−2⋅1, 2⋅2−1⋅(−1), 1⋅1−(−1)⋅2)=(−1,5,3)\vec n=((-1)(-1)-2\cdot1,\ 2\cdot2-1\cdot(-1),\ 1\cdot1-(-1)\cdot2)=(-1,5,3), with ∣n⃗∣=35|\vec n|=\sqrt{35}. Since Π\Pi passes through (1,2,3)(1,2,3) on L1L_1, its equation is −x+5y+3z=−1+10+9=18-x+5y+3z=-1+10+9=18. The distance of the origin is therefore 18/3518/\sqrt{35}. The value 15/3515/\sqrt{35} is the distance between the two skew lines (the point (2,1,0)(2,1,0) of L2L_2 gives ∣−2+5−18∣/35|{-2}+5-18|/\sqrt{35}), which is a different quantity; 3/353/\sqrt{35} is the distance of the origin from the parallel plane through L2L_2.
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Original question written in the JEE Advanced 2020 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake