Let ℝ^3 denote the three-dimensional space. Take two points P=(1,2,3) and Q=(4,2,7) . Let dist(X, Y) denote the distance between two points X and Y in ℝ^3 . Let gathered S=\ X ∈ ℝ^3:(dist(X, P))^2-(dist(X, Q))^2=50\ and

Correct answer: A. There is a triangle whose area is 1 and all of whose vertices are from S . aligned & S=\ X:(XP)^2-(XQ)^2=50\ \& T=\ Y:(YQ)^2-(YP)^2=50\ aligned for finding S X(x, y, z) and for T Y(x, y, z) aligned &((x-1)^2+(y-1)^2+(z-1)^2)-((x-4)^2+(y-2)^2+(z-7)^2)=50 \& S=\ (x, y, z): 6 x+8 z=105\ \& T=\ (x, y, z): 6 x+8 z=5\ aligned Since S and T both are plane ; (1) There exist a triangle in plane S whose area =1 (always) (2) L ~\&~ M lies on plane T , hence line segment joining L \& M will lie on plane T . (3) Distance between S \& ~T d=|(105-5)/(10)|=10 Hence for rectangle of perimeter 48 can exist. (4) For Square There will be infinite such rectangle possible. Hence Answers 1,2,3,4 are correct.

JEE Advanced 2024Paper 1

Exam: JEE Advanced
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Advanced 2024 · Paper 1 · Mathematics · Three Dimensional Geometry

Let ℝ^3 denote the three-dimensional space. Take two points...

Let

\mathbb{R}^3

denote the three-dimensional space. Take two points

P=(1,2,3)

and

Q=(4,2,7)

. Let

\operatorname{dist}(X, Y)

denote the distance between two points

X

and

Y

\mathbb{R}^3

. Let

\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? Let ℝ^3 denote the three-dimensional space. Take two points P=(1,2,3) and Q=(4,2,7) . Let dist(X, Y) denote the

Options

AThere is a triangle whose area is 1 and all of whose vertices are from $S$ .
BThere are two distinct points $L$ and $M$ in $T$ such that each point on the line segment $L M$ is also in $T$ .
CThere are infinitely many rectangles of perimeter 48 , two of whose vertices are from $S$ and the other two vertices are from $T$ .
DThere is a square of perimeter 48 , two of whose vertices are from $S$ and the other two vertices are from $T$ .

Correct answer

A. There is a triangle whose area is 1 and all of whose vertices are from $S$ .

Step-by-step solution

\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 \equiv X(x, y, z)

and for

T \equiv Y(x, y, z)

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

Since

\mathrm{S}

and

\mathrm{T}

both are plane ; (1) There exist a triangle in plane

\mathrm{S}

whose area

(always) (2)

\mathrm{L} ~\&~ \mathrm{M}

lies on plane

\mathrm{T}

, hence line segment joining

\mathrm{L} \& \mathrm{M}

will lie on plane

\mathrm{T}

. (3) Distance between

\mathrm{S} \& \mathrm{~T} \mathrm{d}=\left|\frac{105-5}{10}\right|=10$ Hence for rectangle of perimeter 48 can exist. (4) For Square Let ℝ^3 denote the three-dimensional space. Take two points P=(1,2,3) and Q=(4,2,7) . Let dist(X, Y) denote the

There will be infinite such rectangle possible. Hence Answers 1,2,3,4 are correct.

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