Distance between two points and section formula

Introduction to Three–dimensional Geometry: the distance between two points and the section formula

The familiar formula for the distance between two points in a plane is easily extended from 2D geometry to the following 3D formula.

To prove this equation, construct a rectangular box as shown, where: P₁ and P₂ are opposite vertices.The faces of the box are assumed to be parallel to the coordinate planes.If A(x₂, y₁, z₁) and B(x₂, y₂, z₁) are the vertices of the box, then:

|P₁A| = |x₂ – x₁|

|AB| = |y₂ – y₁|

|BP₂| = |z₂ – z₁|

Triangles P₁BP₂ and P₁AB are right-angled.So, two applications of the Pythagorean Theorem give:

|P₁P₂|² =|P₁B|² + |BP₂|²

|P₁B|² = |P₁A|² + |AB|²

|P₁P₂|² = |P₁A|² + |AB|² + |BP₂|²

= |x₂ – x₁|² + |y₂ – y₁|² + |z₂ – z₁|²

= (x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)²

Therefore,

For example, the distance from the point P(2, –1, 7) to the point Q(1, –3, 5) is:

A section is a point that divides a line segment in a given ratio. To evaluate the section formula in the internal or external division for a line in 3D space we consider the following figure:

For this purpose, consider two points A (x₁, y₁, z₁) and B(x₂, y₂, z₂) and a point P(x, y, z) dividing AB in the ratio m:n. To find point P, we first draw lines AL, PN, and BM that are perpendicular to XY plane such that AL || PN || BM. This means that the points L, M and N lie on the straight line that is made by the intersection of a plane with AL, PN and BM and XY- plane. From point P, a line segment ST is drawn parallel to LM which intersects AL externally at S and it intersects BM internally at T.Since ST ||LM and AL || PN || BM, the quadrilaterals LNPS and NMTP qualify as a parallelogram. Also, ∆ASP is a similar triangle of ∆BTP therefore,

$\frac{m}{n}=\frac{AP}{BP}=\frac{AS}{BT}=\frac{SL-AL}{BM-TM}=\frac{NP-AL}{BM-PN}=\frac{z-z_{1}}{z_{2}-z}$

Making z as the subject of the equation, we rearrange them to obtain:

mz₂ – mz = nz – nz₁

mz₂ + nz₁ = z(n + m)

$z=\frac{mz_{2}+nz_{1}}{m+n}$

Similarly, $x=\frac{mx+nx_{1}}{m+n}$

$y=\frac{my_{2}+ny_{1}}{m+n}$

In conclusion, the coordinates of the point P(x, y, z) that divide the line segment by the points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) in the ratio m:n internally are given by:

$p(x, y,z)=(\frac{mx+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}, \frac{mz_{2}+nz_{1}}{m+n})$

For the case where point P divides the line segment joining the points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) externally in the ratio m:n, then we deduce coordinates of P are by switching n with –n:

$P(x, y,z)=(\frac{mx-nx_{1}}{m-n}, \frac{my_{2}-ny_{1}}{m-n}, \frac{mz_{2}-nz_{1}}{m-n})$

The midpoint M of the line segment PQ with endpointsP(x₁, y₁, z₁ ) and Q(x₂, y₂, z₂ ) is