JEE Advanced 2024 Paper 1 · Q07 · 3D Geometry
Let $\mathbb{R}^3$ denote the three-dimensional space. Take two points $P = (1, 2, 3)$ and $Q = (4, 2, 7)$. Let $\mathrm{dist}(X, Y)$ denote the distance between two points $X$ and $Y$ in $\mathbb{R}^3$. Let $$S = \{X \in \mathbb{R}^3 : (\mathrm{dist}(X, P))^2 - (\mathrm{dist}(X, Q))^2 = 50\}\ \text{and}$$ $$T = \{Y \in \mathbb{R}^3 : (\mathrm{dist}(Y, Q))^2 - (\mathrm{dist}(Y, P))^2 = 50\}.$$ Then which of the following statements is (are) TRUE?
Reveal answer + step-by-step solution
Correct answer:A, B, C, D
Solution
$\mathrm{dist}(X,P)^2-\mathrm{dist}(X,Q)^2 = (X-P)\cdot(X-P)-(X-Q)\cdot(X-Q) = -2X\cdot(P-Q)+|P|^2-|Q|^2$. With $P-Q=(-3,0,-4)$ and $|P|^2-|Q|^2=14-69=-55$, this becomes $6x+8z-55=50\Rightarrow 6x+8z=105$ — a plane. Similarly $T$: $6x+8z=5$ — a plane parallel to $S$. Distance between planes $=|105-5|/\sqrt{36+64}=100/10=10$. (A) Any plane has triangles of arbitrary area. TRUE. (B) Plane $T$ contains line segments. TRUE. (C) Pick any rectangle with two opposite sides of length 10 (perpendicular distance) and 14 (so perimeter $=2(10+14)=48$): infinitely many orientations exist. TRUE. (D) Square needs sides 12 each, but the two parallel sides are forced to be of length equal to the inter-plane distance $10\neq 12$ — wait, square with vertices on parallel planes: two on each. Side $= 12$, with two vertices on each plane, separated by $10$. Possible if $10\leq 12$ (slant). TRUE. All correct: A, B, C, D.
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