Question

Let R³ 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 R3. Let
\(S=\left\{X\in\R^3:(dist(X,P)^2)-(dist(X,Q))^2=50\right\}\ and\)
\(T=\left\{Y\in \R^3:(dist(Y,Q))^2-(dist(Y,P))^2=50\right\}\)
Then which of the following statements is (are) TRUE ?

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

Answer 1

The Correct Option is A, B, C

The given conditions define \( S \) and \( T \) as planes in \( \mathbb{R}^3 \). 

For \( S \), the given equation:

\[ (\text{dist}(X, P))^2 - (\text{dist}(X, Q))^2 = 50 \]

simplifies to:

\[ 6x + 8z = 105. \quad \text{(Equation 1)} \]

For \( T \), the given equation:

\[ (\text{dist}(Y, Q))^2 - (\text{dist}(Y, P))^2 = 50 \]

simplifies to:

\[ 6x + 8z = 5. \quad \text{(Equation 2)} \]

The planes \( S \) and \( T \) are parallel with a constant difference.

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(A) Triangles can be formed by taking any 3 points in the \( XZ \)-plane, with all points lying on \( S \) (\( 6x + 8z = 105 \)). By choosing three appropriate points, a triangle of area 1 can be formed.

(A is True).

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(B) \( T \) is also a plane, and any two distinct points \( L, M \in T \) form a line segment \( LM \) that is completely contained in \( T \), as \( T \) represents a continuous plane.

(B is True).

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(C) The distance between the parallel planes \( S \) and \( T \) is given by:

\[ \text{Distance} = \frac{|105 - 5|}{\sqrt{6^2 + 8^2}} = \frac{100}{10} = 10. \]

Infinitely many rectangles with sides 10 and 14 can be formed, with two vertices on \( S \) and the other two vertices on \( T \). The perimeter of such rectangles is:

\[ 2(10 + 14) = 48. \]

(C is True).

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(D) Given the flexibility of selecting points from \( S \) and \( T \), a square of perimeter 48 can be formed.

(D is True).

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Final Answer:

\[ (A, B, C, D) \]