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JEE Advanced 2022 Paper 2 · Mathematics · Three Dimensional Geometry · medium

A plane contains the line L: (x-1)/(2)=(y)/(-1)=(z-2)/(2) and also passes through the point P(3,4,1). The perpendicular distance of the origin from…

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A plane contains the line L: x−12=y−1=z−22L:\ \dfrac{x-1}{2}=\dfrac{y}{-1}=\dfrac{z-2}{2} and also passes through the point P(3,4,1)P(3,4,1). The perpendicular distance of the origin from this plane is
  1. A. 13185\dfrac{13}{\sqrt{185}}
  2. B. 11185\dfrac{11}{\sqrt{185}}
  3. C. 13150\dfrac{13}{\sqrt{150}}
  4. D. 26185\dfrac{26}{\sqrt{185}}
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Answer: (A) 13185\dfrac{13}{\sqrt{185}}

Take A(1,0,2)A(1,0,2) on LL with direction d⃗=(2,−1,2)\vec d=(2,-1,2). The plane normal is n⃗=d⃗×AP→\vec n=\vec d\times\overrightarrow{AP} with AP→=(2,4,−1)\overrightarrow{AP}=(2,4,-1). Then n⃗=((−1)(−1)−2⋅4, 2⋅2−2⋅(−1), 2⋅4−(−1)⋅2)=(−7,6,10)\vec n=\bigl((-1)(-1)-2\cdot4,\ 2\cdot2-2\cdot(-1),\ 2\cdot4-(-1)\cdot2\bigr)=(-7,6,10). Since n⃗⋅A=−7+0+20=13\vec n\cdot A=-7+0+20=13, the plane is −7x+6y+10z=13-7x+6y+10z=13; check at PP: −21+24+10=13-21+24+10=13. Its distance from the origin is 1349+36+100=13185\dfrac{13}{\sqrt{49+36+100}}=\dfrac{13}{\sqrt{185}}. The option with numerator 2626 doubles the constant, and 150\sqrt{150} comes from a sign slip in computing the normal's length.
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