JEE Advanced 2022 Paper 1 Q11 Mathematics Vectors & 3D Geometry 3D Planes Hard

JEE Advanced 2022 Paper 1 · Q11 · 3D Planes

Let $P_{1}$ and $P_{2}$ be two planes given by $P_{1}:10x+15y+12z-60=0$, $P_{2}:-2x+5y+4z-20=0$. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_{1}$ and $P_{2}$?

  1. A. $\dfrac{x-1}{0}=\dfrac{y-1}{0}=\dfrac{z-1}{5}$
  2. B. $\dfrac{x-6}{-5}=\dfrac{y}{2}=\dfrac{z}{3}$
  3. C. $\dfrac{x}{-2}=\dfrac{y-4}{5}=\dfrac{z}{4}$
  4. D. $\dfrac{x}{1}=\dfrac{y-4}{-2}=\dfrac{z}{3}$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, D

Solution

Edges of such a tetrahedron either lie along $P_{1}\cap P_{2}$, or lie on exactly one plane, or join non-shared vertices and meet each plane at a single point. Line of intersection direction: $\vec{n_{1}}\times\vec{n_{2}}=(10,15,12)\times(-2,5,4)\propto(0,-4,5)$. (A) dir $(0,0,5)$ not parallel to either plane; passes through $(1,1,1)$ which is on neither plane, crosses both — valid edge. (B) dir $(-5,2,3)$; not in $P_{1}$ (substitute normal: $10(-5)+15(2)+12(3)=16\ne 0$); not in $P_{2}$ either; not parallel to either plane, so meets both — valid. (C) dir $(-2,5,4)=\vec{n_{2}}$ — perpendicular to $P_{2}$. A line perpendicular to one face cannot serve as an edge of that tetrahedron geometrically. NOT valid. (D) dir $(1,-2,3)$; not parallel to either plane; intersects each at one point — valid. Answer: A, B, D.

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