x^2 + y^2 - 2y + 1 + z^2 - (x^2 - 2x + 1 + y^2 + z^2) = 0
Title: Simplifying and Interpreting the 3D Equation: A Comprehensive Guide to x² + y² − 2y + 1 + z² − (x² − 2x + 1 + y² + z²) = 0
Meta Description:Explore the simplification and geometric meaning of the 3D equation x² + y² − 2y + 1 + z² − (x² − 2x + 1 + y² + z²) = 0. Discover how this equation describes a point in space and how to rewrite it in standard form.
Introduction
Mathematical equations often encode rich geometric information, especially in three dimensions. Today, we analyze and simplify a key equation:
[x^2 + y^2 - 2y + 1 + z^2 - (x^2 - 2x + 1 + y^2 + z^2) = 0]
Under the hood, this equation represents a point in space—specifically, it reduces to a single coordinate condition, revealing a specific location in 3D geometry. Let’s break this down step by step.
Step 1: Expand and Simplify the Expression
Start by expanding both sides of the equation. Note that the expression includes a parenthetical term:[-(x^2 - 2x + 1 + y^2 + z^2)]
Distribute the negative sign:
[x^2 + y^2 - 2y + 1 + z^2 - x^2 + 2x - 1 - y^2 - z^2 = 0]
Now combine like terms:
- (x^2 - x^2 = 0)- (y^2 - y^2 = 0)- (z^2 - z^2 = 0)- (-2y) remains- (+1 - 1 = 0)- (+2x) remains
After cancellation, the entire left-hand side reduces to:
[-2y + 2x = 0]
So:
[2x - 2y = 0 \quad \Rightarrow \quad x = y]
Step 2: Interpretation in 3D Space
At first glance, this appears degenerate—a 2D plane (x = y) extended along (z). However, note that the variables (z) and higher-degree terms canceled out completely, leaving only the condition (x = y), independent of (z) and (y).
This means the "equation" describes an infinite vertical line* in 3D space, but restrict further: since no specific bound is placed on (x) and (y), and (z) appears symmetrically and cancels, what does this represent?
Actually, re-examining carefully: the original expression contains no (z)^3 or nonlinear terms—but crucially, there is no constraint on (z). However, in the simplification, all (z)-dependent terms canceled identically, and no dependency remains except influencing a line condition in (x) and (y).
But let’s be precise: after full simplification:
[2x - 2y = 0 \quad \Rightarrow \quad x = y \quad \ ext{for all } z]
This is a plane extending vertically (along (z)) where every point lies on the line (x = y) in 3D space.
But wait—is it a line or a plane? Because (z) is arbitrary, the set of solutions is the entire line (x = y), parallel to the (z)-axis, everywhere in space.
However, upon deeper inspection, let’s verify that no interference from (z) exists and that reduction is valid.
Looking back:
Original:[x^2 + y^2 - 2y + 1 + z^2 - (x^2 - 2x + 1 + y^2 + z^2)]
Group terms:
- (x^2 - x^2 = 0)- (y^2 - y^2 = 0)- (z^2 - z^2 = 0)- (-2y + 2x)- (1 - 1 = 0)
So all variable terms vanish except (2x - 2y)
Thus, the equation reduces identically to:
[2x - 2y = 0 \quad \Rightarrow \quad x = y]
This defines a plane? No—an entire vertical line in 3D space only if (z) were constrained—but here (z) is free.Actually, since (z) cancels out and no condition restricts it, (z \in \mathbb{R}), so the solution is the set of all points ((x, y, z)) such that (x = y)—this is a plane in 3D space: a vertical plane slicing through the origin along (x = y).
Wait—clarify: such a surface is actually a plane, not a line. Why? Because fixing (x = y) over all (z) forms an infinite plane with normal vector perpendicular to ((1, -1, 0)). For example, any point ((t, t, s)) satisfies (x = y), so the solution set is the plane:
[\boxed{x = y}]
This is a 2D plane in 3D space, not a line—because (x) and (y) are dynamically linked, while (z) remains free.
But wait: in Cartesian geometry, the surface defined by (x = y) is a plane that cuts through the origin at a 45° angle in the (xy)-plane and extends infinitely in all directions, including all values of (z).
Step 3: Is This Just a Plane? Clarifying the Geometry
While it looks like a line initially due to reduction to (x = y), the fact that (z) is unrestricted changes everything:
- If the equation imposed a bound on (z), it might limit to a line.- But here, no restriction on (z) remains after simplification.
Therefore, the solution set is uncountably infinite, forming a plane in 3D space defined by (x = y), with (z) arbitrary.
This is equivalent to the affine plane in (\mathbb{R}^3):
[{(x, y, z) \mid x = y}]
Geometrically, this plane:
- Passes through the origin.- Is orthogonal to the vector ((1, -1, 0)).- Intersects the (xy)-plane along the line (x = y), and extends infinitely in the (z)-direction.
Step 4: Alternative Approach — Rewrite in Standard Form
Start over:
Given:[x^2 + y^2 - 2y + 1 + z^2 = x^2 - 2x + 1 + y^2 + z^2]
Subtract right-hand side from left:
[(x^2 + y^2 - 2y + 1 + z^2) - (x^2 - 2x + 1 + y^2 + z^2) = 0]
Simplify:
[y^2 - y^2 + z^2 - z^2 + x^2 - x^2 + 2x - 2y + 1 - 1 = 0]
[2x - 2y = 0]
[x = y]
Confirmed: the equation reduces exactly to (x = y), a plane.
Step 5: Graphical and Algebraic Interpretation
- Graphically: All points ((x, y, z)) where the (x)- and (y)-coordinates are equal form a vertical plane slicing through space like a Milwaukee sheet inclining at 45° in the (xy)-plane.- Algebraically: The original expression was balanced only by setting (x = y), with no constraints on (z), proving that (z) is not fixed.
Why Is This Useful?
Understanding such simplified forms helps in:
- 3D visualization in geometry and physics.- Solving systems involving multiple equations.- Optimization over spatial domains, where constraints may reduce dimensionality.
Conclusion
The equation:
[x^2 + y^2 - 2y + 1 + z^2 - (x^2 - 2x + 1 + y^2 + z^2) = 0]
simplifies to:
[x = y]
which defines a vertical plane in three-dimensional space, independent of (z). This means all points ((x, y, z)) where (x = y) satisfy the equation—this is a linear subspace (a plane)—not just a line or point.
So while the form looks like a constraint yielding a degenerate case, careful simplification confirms it’s a genuine plane: one of the most fundamental 3D geometric objects.
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