JEE Advanced 2022 — Paper 1 — MATH-II solution

JEE Advanced2022Paper 1MATH-II

Let $P_{1}$ and $P_{2}$ be two planes given by

$P_{1} : 10 x + 15 y + 12 z - 60 = 0,$
$P_{2} : - 2 x + 5 y + 4 z - 20 = 0.$

Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_{1}$ and $P_{2}$ ?

A
$\frac{x - 1}{0} = \frac{y - 1}{0} = \frac{z - 1}{5}$
B
$\frac{x - 6}{- 5} = \frac{y}{2} = \frac{z}{3}$
C
$\frac{x}{- 2} = \frac{y - 4}{5} = \frac{z}{4}$
D
$\frac{x}{1} = \frac{y - 4}{- 2} = \frac{z}{3}$

Solution

A line can be an edge of a tetrahedron with two faces on $P_{1}$ and $P_{2}$ provided the line either lies in one of the planes and intersects the other, or is the common edge (line of intersection) of the two planes, or meets each plane in a separate point.

Equivalently, the line should meet the surface $S = P_{1} \cdot P_{2} = 0$ in at least two distinct points (or in infinitely many, lying in $P_{1} \cap P_{2}$ ). Substitute a parametric point of each line into $S$ and analyse the roots in the parameter $λ$ .

(A) Point $(1, 1, 5 λ + 1)$ . Substituting into $S$ gives $(60 λ - 23) (20 λ - 17) = 0$ , with two real roots $λ = \frac{23}{60}, \frac{17}{20}$ . The line meets the two planes at distinct points $\Rightarrow$ admissible edge.

(B) Point $(- 5 λ + 6, 2 λ, 3 λ)$ . Substituting gives $(16 λ) (32 λ - 32) = 0$ , with roots $λ = 0$ and $λ = 1$ $\Rightarrow$ admissible edge.

(C) Point $(- 2 λ, 5 λ + 4, 4 λ)$ . Substituting gives $(103 λ) (45 λ) = 0$ , both factors vanish only at $λ = 0$ . Only one intersection point $\Rightarrow$ NOT an edge.

(D) Point $(λ, - 2 λ + 4, 3 λ)$ . Substituting gives $(16 λ) (0) = 0$ , which holds for every $λ$ . The line lies entirely on $P_{2}$ $\Rightarrow$ admissible edge.

$∴$ Options (A), (B) and (D) are correct.

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