Verified Solution • 2025–26 Session moderate

QJEEAdv24Madebd: JEE Advanced 2024 Mathematics - Class 12 JEEAdv24Madebd Solution: Let denote the three-dimensional space. Take two points and .…

Let R3\mathbb{R}^3 denote the three-dimensional space. Take two points P=(1,2,3)P=(1,2,3) and Q=(4,2,7)Q=(4,2,7) . Let dist⁡(X,Y)\operatorname{dist}(X, Y) denote the distance between two points XX and YY in R3\mathbb{R}^3 . Let

S={X∈R3:(dist⁡(X,P))2−(dist⁡(X,Q))2=50} and T={Y∈R3:(dist⁡(Y,Q))2−(dist⁡(Y,P))2=50}.\begin{gathered} S=\left\{X \in \mathbb{R}^3:(\operatorname{dist}(X, P))^2-(\operatorname{dist}(X, Q))^2=50\right\} \text { and } \\ T=\left\{Y \in \mathbb{R}^3:(\operatorname{dist}(Y, Q))^2-(\operatorname{dist}(Y, P))^2=50\right\} . \end{gathered}

Then which of the following statements is (are) TRUE?

Multiple Choice Options

  • A There is a triangle whose area is 1 and all of whose vertices are from SS .
  • B There are two distinct points LL and MM in TT such that each point on the line segment LML M is also in TT .
  • C There are infinitely many rectangles of perimeter 48 , two of whose vertices are from SS and the other two vertices are from TT .
  • D There is a square of perimeter 48 , two of whose vertices are from SS and the other two vertices are from TT .

Step-by-Step Verified Solution

amp;S={X:(XP)2−(XQ)2=50}amp;T={Y:(YQ)2−(YP)2=50}\begin{aligned}& \mathrm{S}=\left\{\mathrm{X}:(\mathrm{XP})^2-(\mathrm{XQ})^2=50\right\} \\& \mathrm{T}=\left\{\mathrm{Y}:(\mathrm{YQ})^2-(\mathrm{YP})^2=50\right\}\end{aligned}

for finding S≡X(x,y,z)S \equiv X(x, y, z) and for T≡Y(x,y,z)T \equiv Y(x, y, z)

amp;((x−1)2+(y−1)2+(z−1)2)−((x−4)2+(y−2)2+(z−7)2)=50amp;⇒S={(x,y,z):6x+8z=105}amp;T={(x,y,z):6x+8z=5}\begin{aligned}& \left((x-1)^2+(y-1)^2+(z-1)^2\right)-\left((x-4)^2+(y-2)^2+(z-7)^2\right)=50 \\& \Rightarrow \quad S=\{(x, y, z): 6 x+8 z=105\} \\& \quad \quad T=\{(x, y, z): 6 x+8 z=5\}\end{aligned}

Since S and T both are plane;

(A) There exist a triangle in plane S whose area =1=1 (always)

(B) L & M lies on plane T, hence line segment joining L & M will lie on plane T.

(C) Distance between S\mathrm{S} & T\mathrm{T}

d=∣105−510∣=10\mathrm{d}=\left|\frac{105-5}{10}\right|=10

Hence for rectangle of perimeter 48 can exist.

JEE Advanced 2024 Paper 1 Online Mathematics - 3D Geometry Question 3 English Explanation 1

There will be infinite such rectangle possible.

(D) For square

JEE Advanced 2024 Paper 1 Online Mathematics - 3D Geometry Question 3 English Explanation 2

Hence Answers A, B, C, D are correct.

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring