1.9 Distance between points in three-dimensional Euclidean space
You saw in Section 1.5 that the distance between two points (x1, y1) and (x2, y2) in the plane is given by
We can establish a similar formula for the distance between two points in 3, as follows.
Let P (x1, y1, z1) and Q(x2, y2, z2) be two points in 3. Let M be the foot of the perpendicular from Q to the plane through P that is parallel to the (x, y)-plane; then M has coordinates (x2, y2, z1). Next, let N be the point in this plane with coordinates (x1, y2, z1).
Then the triangles PQM and PMN are both right-angled triangles, with right angles at M and N, respectively, as shown below.
The length of PN is |y2 − y1|, the length of NM is |x2 − x1|, and the length of MQ is |z2 − z1|. It follows from Pythagoras' Theorem that
PM2 = NM2 + PN2,
PM2 = (x2 − x1)2 + (y2 − y1)2.
Note: This part of the discussion is similar to the derivation of the Distance Formula in 2 (see Section 1.5).
Using Pythagoras' Theorem again, we obtain
PQ2 = PM2 + MQ2
Distance Formula in three-dimensional Euclidean space
The distance between two points (x1, y1, z1) and (x2, y2, z2) in 3 is
For example, it follows from this formula that the distance between the points (1, 2, 3) and (4, −2, 15) is
For each of the following pairs of points in 3, find the distance between them.
(a) (1, 1, 1) and (4, 1, −3).
(b) (1, 2, 3) and (3, 0, 3).
We use the formula for the distance between two points in 3 This gives the following distances.