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

To solve the problem, we need to analyze the sets \( S \) and \( T \) defined in \( \mathbb{R}^3 \) and verify the given statements.

1. Understanding the Sets \( S \) and \( T \):
Given points \( P = (1, 2, 3) \) and \( Q = (4, 2, 7) \), the distance between two points \( X = (x, y, z) \) and \( Y \) in \( \mathbb{R}^3 \) is given by:

\( \text{dist}(X, Y) = \sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2} \), where \( Y = (a, b, c) \).

The sets \( S \) and \( T \) are defined as:

\( S = \{ X \in \mathbb{R}^3 : \text{dist}(X, P)^2 - \text{dist}(X, Q)^2 = 50 \} \)
\( T = \{ Y \in \mathbb{R}^3 : \text{dist}(Y, Q)^2 - \text{dist}(Y, P)^2 = 50 \} \).

2. Simplifying the Conditions for \( S \) and \( T \):
For \( S \), expand the distance expressions:

\( (x - 1)^2 + (y - 2)^2 + (z - 3)^2 - [(x - 4)^2 + (y - 2)^2 + (z - 7)^2] = 50 \).

Simplify:

\( (x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 6z + 9) - (x^2 - 8x + 16) - (y^2 - 4y + 4) - (z^2 - 14z + 49) = 50 \).

Cancel terms and combine like terms:

\( 6x + 8z - 55 = 50 \)
\( 6x + 8z = 105 \).

Thus, \( S \) is the plane \( 6x + 8z = 105 \). Similarly, for \( T \):

\( (x - 4)^2 + (y - 2)^2 + (z - 7)^2 - [(x - 1)^2 + (y - 2)^2 + (z - 3)^2] = 50 \).

Simplify:

\( -6x - 8z + 55 = 50 \)
\( -6x - 8z = -5 \)
\( 6x + 8z = 5 \).

Thus, \( T \) is the plane \( 6x + 8z = 5 \).

3. Analyzing the Statements:
Statement 1: "There is a triangle whose area is 1 and all of whose vertices are from \( S \)."
Since \( S \) is a plane, we can find three non-collinear points in \( S \) forming a triangle. The area of a triangle can be adjusted by scaling the coordinates, so it is possible to have a triangle with area 1. Thus, this statement is TRUE.

Statement 2: "There are two distinct points \( L \) and \( M \) in \( T \) such that each point on the line segment \( LM \) is also in \( T \)."
Since \( T \) is a plane, any two points in \( T \) will have their entire line segment lying in \( T \). Thus, this statement is TRUE.

Statement 3: "There are infinitely many rectangles of perimeter 48, two of whose vertices are from \( S \) and the other two vertices are from \( T \)."
The distance between the parallel planes \( S \) and \( T \) can be calculated. The normal vector is \( (6, 0, 8) \), and the distance \( d \) between the planes is:

\( d = \frac{|105 - 5|}{\sqrt{6^2 + 8^2}} = \frac{100}{10} = 10 \).

For a rectangle with two vertices in \( S \) and two in \( T \), the side lengths must satisfy \( 2(a + b) = 48 \), i.e., \( a + b = 24 \). The distance between \( S \) and \( T \) fixes one side (say \( b = 10 \)), so \( a = 14 \). There are infinitely many such rectangles, so this statement is TRUE.

Statement 4: "There is a square of perimeter 48, two of whose vertices are from \( S \) and the other two vertices are from \( T \)."
For a square, all sides must be equal. Here, one pair of sides is 10 (distance between \( S \) and \( T \)), and the other sides must also be 10. However, the perimeter would be \( 4 \times 10 = 40 \), not 48. Thus, no such square exists, and this statement is FALSE.

Final Answer:
The statements that are TRUE are \(\boxed{1}\), \(\boxed{2}\), and \(\boxed{3}\).

Answer 2

The problem involves determining the nature of the sets \( S \) and \( T \) based on conditions involving the distances from points \( P = (1, 2, 3) \) and \( Q = (4, 2, 7) \). We compute the sets where the distance criteria are: 
\((dist(X,P))^2 - (dist(X,Q))^2 = 50\) for set \( S \) and \((dist(Y,Q))^2 - (dist(Y,P))^2 = 50\) for set \( T \).

We first establish the equations of these sets. The squared distance from a point \( X = (x_1, x_2, x_3) \) to \( P \) is \[ (x_1-1)^2 + (x_2-2)^2 + (x_3-3)^2 \]. To \( Q \), it is \[ (x_1-4)^2 + (x_2-2)^2 + (x_3-7)^2 \]. For set \( S \), the condition leads to:
 

\[((x_1-1)^2 + (x_2-2)^2 + (x_3-3)^2) - ((x_1-4)^2 + (x_2-2)^2 + (x_3-7)^2) = 50\]

Simplifying gives the plane equation: 
\[-6x_1 +4x_3 = 50\].
Similarly, for set \( T \), the condition gives:
\[((x_1-4)^2 + (x_2-2)^2 + (x_3-7)^2) - ((x_1-1)^2 + (x_2-2)^2 + (x_3-3)^2) = 50 \]
Simplifying gives another plane equation: 
\[6x_1-4x_3 = 50\].

The set \( S \) and \( T \) represent parallel planes in \( \mathbb{R}^3 \). Each set is a plane without dependence on \( x_2 \), implying infinite parallel lines.

Examining the statements:

• "There is a triangle whose area is 1 and all of whose vertices are from S." There is infinite freedom to select vertices on plane \( S \) such that the triangle area is 1 using the general geometric configurations, making this statement true.
• "There are two distinct points L and M in T such that each point on the line segment LM is also in T." Since \( T \) is a plane, any points \( L \) and \( M \) lie on \( T \), and so does the line segment connecting them. This is true.
• "There are infinitely many rectangles of perimeter 48, two of whose vertices are from S and the other two vertices are from T." This is possible due to the parallel nature of the planes, setting appropriate coordinates minimizes any dependency on \( x_2 \) allowing flexible points choice to design a rectangle of perimeter 48. Thus, this is true.
• "There is a square of perimeter 48, two of whose vertices are from S and the other two vertices are from T." Squares are restricted by having equal sides, constraining options too narrowly between planes, making it more difficult to achieve such a configuration under the given conditions. This is false.