Distance Between Two Points

Distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in space. Direct extension of the 2D formula using Pythagoras twice.

Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be two points in space.

Step 1: Drop a perpendicular from $P$ to the xy-plane, meeting it at $A(x_1, y_1, 0)$. Similarly $B(x_2, y_2, 0)$ for $Q$.

Step 2: In the xy-plane, $AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

Step 3: $PQ$ is the hypotenuse of a right triangle with legs $AB$ (horizontal) and $z_2-z_1$ (vertical):

$$PQ = \sqrt{AB^2 + (z_2-z_1)^2} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

Distance from origin: For $P(x, y, z)$:

$$OP = \sqrt{x^2+y^2+z^2}$$

Collinearity test: Three points $A$, $B$, $C$ are collinear iff $AB + BC = AC$ (one distance equals the sum of the other two).