Questions: Distance Formula and Metric in 3D Space

Question 1

What is the distance between the points (1, 2, 3) and (4, 6, 3)?

• A. 5, since √[(4−1)² + (6−2)² + (3−3)²] = √25
• B. √7, since √(3 + 4 + 0)
• C. √34, since √(9 + 16 + 9)
• D. 7, since (4−1) + (6−2) + (3−3)

Answer: (A)

Distance = √[(4−1)² + (6−2)² + (3−3)²] = √[9 + 16 + 0] = √25 = 5. Since the z-coordinates are equal, the z-term contributes zero and this reduces to the 2D distance in the xy-plane. Option C incorrectly uses (3−3)² = 9 instead of 0; option B adds differences without squaring; option D omits the square root entirely.

Question 2

Which equation describes all points (x, y, z) at distance 3 from the point (1, 2, 4)?

• A. (x−1) + (y−2) + (z−4) = 3
• B. (x−1)² + (y−2)² + (z−4)² = 9
• C. √[(x−1)² + (y−2)² + (z−4)²] = 9
• D. (x+1)² + (y+2)² + (z+4)² = 3

Answer: (B)

Set the distance formula equal to r = 3 and square both sides: (x−1)² + (y−2)² + (z−4)² = 9. This is the equation of a sphere of radius 3 centered at (1, 2, 4). Option C has 9 under the root instead of 3 (the formula would give distance = √9 = 3, but it's written inconsistently); option D has the wrong signs on the center coordinates and the wrong exponent on r; option A omits squaring the differences.

Question 3

The 3D distance formula is derived from a single application of the Pythagorean theorem, extended to accommodate a third coordinate.

• T. True
• F. False

Answer: False

Two applications are required. First, compute the horizontal distance in the xy-plane: d_xy = √[(x₂−x₁)² + (y₂−y₁)²]. Then treat d_xy and |z₂−z₁| as legs of a new right triangle and apply Pythagoras again: distance = √[d_xy² + (z₂−z₁)²] = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]. This two-step structure is why the formula works: the three axes are mutually perpendicular, so each coordinate difference contributes independently.

Question 4

The Euclidean distance between two points gives the length of the shortest possible path between them through 3D space.

• T. True
• F. False

Answer: True

In Euclidean space, the straight-line segment is always the shortest path between two points — this follows from the triangle inequality, which the Euclidean metric satisfies. The formula gives exactly this straight-line length. It is a geometric property of space itself, not a navigational calculation. This is why 'Euclidean metric' is the standard meaning of 'distance' in 3D space, and why it generalizes naturally to higher dimensions.

Question 5

Explain why the 3D distance formula adds a (z₂−z₁)² term rather than simply using the 2D formula with an additive z-correction.

Model answer: Because the three coordinate axes are mutually perpendicular, displacements in each direction contribute independently and orthogonally to total distance. The 2D formula gives the horizontal distance d_xy in the xy-plane. The actual 3D path from one point to the other is the hypotenuse of a right triangle whose legs are d_xy and |z₂−z₁|. Applying the Pythagorean theorem to this triangle gives distance² = d_xy² + (z₂−z₁)² = (x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)². The squaring and square-rooting follow directly from this second Pythagorean application.

The key geometric insight is that the three Cartesian axes are orthogonal — they don't cross-couple. This is what allows the formula to be a simple sum of squared differences. In a non-Cartesian system where axes are not perpendicular, you would need cross-terms (as in the law of cosines). The elegant form of the Euclidean metric reflects the orthogonality of the coordinate system.