Let R³ denote the three dimensional space. Take two points P = (1, 2, 3) and Q = (4, 2, 7). Let dist

Q: Wait, why do the squared terms cancel? I thought distance formulas always have square roots.

That is a fantastic observation! The problem gives us the *squared* distance, $(\text{dist}(X, P))^2$. This is a huge hint! By squaring the distance, we eliminate the square root, which transforms a complex radical equation into a simple linear one. Always look for these structural gifts in JEE problems.

Q: How can a square of side 12 fit between planes that are only 10 units apart?

Think of the square as a rigid sheet. If you place it flat, it won't fit. But if you tilt it, the perpendicular projection of the square onto the normal vector of the planes is $s \cos \theta$. As long as the side length $s$ is greater than or equal to the distance $d$ between the planes, you can always find an angle $\theta$ such that $s \cos \theta = d$. So, yes, it fits perfectly!

Q: Why are $S$ and $T$ planes? I thought distance equations usually defined spheres.

You are right that $dist(X, P) = k$ defines a sphere. However, the condition $(\text{dist}(X, P))^2 - (\text{dist}(X, Q))^2 = k$ is the definition of a *hyperboloid* in general, but because the coefficients of the quadratic terms cancel out here, it collapses into a plane. It is a special case where the locus is a flat surface!

Question

Let $\mathbb{R}^{3}$ denote the three dimensional space. Take two points $P = (1, 2, 3)$ and $Q = (4, 2, 7)$. Let $	ext{dist}(X, Y)$ denote the distance between two points $X$ and $Y$ in $\mathbb{R}^{3}$. Let $S = \{X \in \mathbb{R}^{3} : (	ext{dist}(X, P))^{2} - (	ext{dist}(X, Q))^{2} = 50\}$ and $T = \{Y \in \mathbb{R}^{3} : (	ext{dist}(Y, Q))^{2} - (	ext{dist}(Y, P))^{2} = 50\}$. Then which of the following statements is(are) TRUE?

Options:

(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 segments 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.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have two fixed points, $P = (1, 2, 3)$ and $Q = (4, 2, 7)$.

You are asked to find the set of all points $X$ that satisfy a specific condition: the difference of the squares of their distances from $P$ and $Q$ must be exactly $50$.

When we write out the condition $(	ext{dist}(X, P))^2 - (	ext{dist}(X, Q))^2 = 50$, we are essentially looking at:

$$ (x-1)^2 + (y-2)^2 + (z-3)^2 - ((x-4)^2 + (y-2)^2 + (z-7)^2) = 50 $$

The magic happens immediately: the $x^2$, $y^2$, and $z^2$ terms on both sides are identical and cancel out completely. We are left with a beautiful, simple linear equation:

$$ 6x + 8z = 105 $$

This is the equation of a plane.

### The Parallel Universe of Planes

Now, consider locus $T$ defined by the condition $(	ext{dist}(Y, Q))^2 - (	ext{dist}(Y, P))^2 = 50$. This is simply the negative of our previous condition.

Following the same algebraic steps, we arrive at:

$$ 6x + 8z = 5 $$

We now have two planes, $S: 6x + 8z = 105$ and $T: 6x + 8z = 5$. Their normal vectors are identical, meaning they are perfectly parallel.

The distance between these two planes is calculated using the standard formula:

$$ d = \frac{|105 - 5|}{\sqrt{6^2 + 0^2 + 8^2}} = \frac{100}{10} = 10 $$

We have established that the planes are exactly **$10$ units apart**.

### Evaluating the Geometric Reality

Now, let us tackle the implications for these geometric loci:

**Option A:** Since $S$ is an infinite plane, we can pick any three non-collinear points to form a triangle of any area we desire. Therefore, a triangle of area $1$ exists on $S$.

**Option B:** A plane is a convex set; any line segment connecting two points on it must lie entirely within it. Thus, a line segment $LM$ on $T$ is entirely contained within $T$.

**Option C:** We consider a rectangle of perimeter $48$ spanning across $S$ and $T$. If we set the width to the distance between the planes, $w = 10$, then the perimeter $2(l + 10) = 48$ gives us $l = 14$. Since $14 > 0$, such a rectangle exists.

**Option D:** For a square of perimeter $48$, each side is $s = 12$. Since $12 > 10$, we can tilt the square diagonally between the planes so that its projection fits the $10$-unit gap.

**Conclusion:** All four statements are true. This problem teaches us that behind every complex-looking equation lies a simple, elegant geometric truth.

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Visualized Solution

Defining Locus $S$

Distance Formula for $S$

Simplifying Locus $S$

Geometric Meaning of $S$

Defining Locus $T$

Simplifying Locus $T$

Parallel Planes $S$ and $T$

Distance Between Planes

Option A: Triangle on $S$

Option B: Line Segment on $T$

Option C: Rectangle Between $S$ and $T$

Option D: Square Between $S$ and $T$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Parallel Universe of Planes

Evaluating the Geometric Reality

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Visualized Solution

Defining Locus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Simplifying Locus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Geometric Meaning of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Defining Locus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Simplifying Locus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Parallel Planes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T