Distance between two points in 3D

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ in the plane is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ , just the Pythagorean theorem. In space, the same principle applies, with one extra coordinate.

The formula

For two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$: $\boxed{|PQ| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}.}$

Derivation

Project $P$ and $Q$ down to the $xy$-plane: their projections are $P'(x_1, y_1, 0)$ and $Q'(x_2, y_2, 0)$. The distance $|P'Q'|$ in the plane is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Now consider the right triangle with one leg along $|P'Q'|$ in the $xy$-plane and the other leg vertical of length $|z_2 - z_1|$. The hypotenuse $|PQ|$ satisfies $|PQ|^2 = |P'Q'|^2 + |z_2 - z_1|^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2. \qed$

Distance from origin

$|OP| = \sqrt{x_1^2 + y_1^2 + z_1^2}$.

Worked examples

Example 1. Distance between $(1, 2, 3)$ and $(4, 6, 8)$.

$\sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$.

Example 2. Distance from origin to $(1, -2, 2)$.

$\sqrt{1 + 4 + 4} = 3$.

Example 3. Show that $(0, 0, 0), (3, 4, 0), (0, 4, 3)$ form an isosceles triangle.

Sides: $AB = \sqrt{9 + 16 + 0} = 5$. $AC = \sqrt{0 + 16 + 9} = 5$. $BC = \sqrt{9 + 0 + 9} = \sqrt{18}$. $AB = AC$, so isosceles.

Example 4. Find the point on the $x$-axis equidistant from $(2, 3, -1)$ and $(4, -1, 5)$.

Let the point be $(a, 0, 0)$.

$\sqrt{(a-2)^2 + 9 + 1} = \sqrt{(a-4)^2 + 1 + 25}$.

Square: $(a-2)^2 + 10 = (a-4)^2 + 26 \Rightarrow a^2 - 4a + 4 + 10 = a^2 - 8a + 16 + 26 \Rightarrow 4 a = 28 \Rightarrow a = 7$.

So $(7, 0, 0)$.

Example 5. Show that $(1, 2, 3), (-1, 0, 4), (4, 1, 2)$ form a right triangle.

$AB^2 = 4 + 4 + 1 = 9$. $AC^2 = 9 + 1 + 1 = 11$. $BC^2 = 25 + 1 + 4 = 30$.

Check: $AB^2 + AC^2 = 9 + 11 = 20 \ne 30$. Try other combinations: $AB^2 + BC^2 = 39 \ne 11$. $AC^2 + BC^2 = 41 \ne 9$. None equal. So not a right triangle. (I'd correct the example with proper integers.)

Let's use a cleaner case: $(0, 0, 0), (1, 0, 0), (0, 1, 0)$. $AB = 1, AC = 1, BC = \sqrt{2}$. $1 + 1 = 2 = BC^2$. Right triangle at $A$.

Try it yourself

1. Distance between $(1, 0, 1)$ and $(0, 1, 0)$.
2. Distance between $(-3, 4, 5)$ and $(2, -1, 0)$.
3. Distance from $(0, 0, 0)$ to $(2, -2, 1)$.
4. Find $a$ such that the distance between $(a, 1, 2)$ and $(3, 1, 5)$ is $5$.
5. Show $(1, 2, 3), (-2, 3, 5), (7, 0, -1)$ are collinear.
6. Find a point on the $y$-axis equidistant from $(1, 2, 3)$ and $(2, 1, 1)$.
7. Find a point on the $z$-axis at distance $7$ from $(2, 3, -1)$.
8. Are $(0, 7, -10), (1, 6, -6), (4, 9, -6)$ vertices of a right triangle?
9. The vertices of a triangle are $(2, 1, 0), (-1, -1, 2), (0, 1, 3)$. Find its perimeter.
10. Find the equidistant locus of $(0, 0, 0)$ and $(2, 0, 0)$.
11. Find the locus of a point that is at distance $3$ from $(1, 2, -1)$.
12. The point $P(x, y, z)$ is equidistant from $(0, 0, 0)$ and $(1, 1, 1)$. Find the equation describing this.

Pitfalls / Tricks

• Don't forget the third term in the square root.
• Distance is always non-negative; take the principal square root.
• For "equidistant" problems, equate squared distances , avoids square roots until you must.
• Insight. The distance formula is the same Pythagorean theorem we used in 2D, applied twice , once in the $xy$-plane, once between that and the vertical lift.