Distance between two points

The distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is the 2D formula with one more square under the root:

• PQ = √((x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)²).
• Why: PQ is the long diagonal of a box with edges equal to the three differences. Pythagoras on the floor of the box gives the floor diagonal; Pythagoras again with the height gives PQ.

Examples:

• (2, –1, 3) to (4, 1, 2): differences 2, 2, –1; 4 + 4 + 1 = 9, so the distance is 3.
• From the origin: OP = √(x² + y² + z²). For (1, 2, 2): √9 = 3.

A tidy habit: write the two x's, subtract and square; then the two y's; then the two z's. Squaring removes any minus sign, so the order of the points does not matter.

Finding a point. Which points on the y-axis are 3 units from (2, 1, 1)? A point on the y-axis is (0, y, 0).

• Square both sides to avoid the root: 4 + (y – 1)² + 1 = 9.
• (y – 1)² = 4, so y – 1 = ±2 and y = 3 or –1.
• The points are (0, 3, 0) and (0, –1, 0).

Three points in a line (collinear). They are collinear when the two shorter distances add up to the longest.

• A(1, 2, 3), B(2, 4, 5), C(4, 8, 9): AB = √(1 + 4 + 4) = 3, BC = √(4 + 16 + 16) = 6, AC = √(9 + 36 + 36) = 9.
• 3 + 6 = 9, so the points are collinear, with B between A and C.

Right-angled and isosceles. A(1, 1, 1), B(3, 1, 1), C(1, 3, 1): AB = 2, AC = 2, BC = √8.

• Two equal sides: isosceles.
• AB² + AC² = 4 + 4 = 8 = BC²: by Pythagoras' converse, right-angled at A.

Four points. Opposite sides equal makes a parallelogram; if the two diagonals are also equal, it is a rectangle.