Vectors and conics

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# 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 |y2y1|, the length of NM is |x2x1|, and the length of MQ is |z2z1|. It follows from Pythagoras' Theorem that

PM2 = NM2 + PN2,

so

PM2 = (x2x1)2 + (y2y1)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

so

that is,

## 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

## Example 11

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.

M208_1