# How to Find Rectangular Coordinate System in Space

The coordinate system defines the position of a vector. When we consider the rectangular coordinates in space, we refer to the three-dimensional space we live in. To demonstrate the position of a vector, we select a point as the origin and represent by the point ‘O’. The distance of any vector is measured form this standard point O. In this article, we will learn how to find rectangular coordinate system in space.

## Coordinate Axes and Coordinate Planes in 3D Space

Let us consider three planes intersecting at a point O such that these three planes are mutually perpendicular to each other. These three planes intersect along the lines X′OX, Y′OY, and Z′OZ, called the x, y, and z-axes, respectively. These lines are mutually perpendicular to each other. These lines form the rectangular coordinate system. The plane XOY is the XY plane. YOZ is the YZ plane and ZOX is the ZX plane. These are called the coordinate planes. The coordinates of the origin O are (0,0,0).

### Coordinates of a Point in Space

Consider a point P in space, we drop a perpendicular PM on the XY-plane with M as the foot of this perpendicular. consider OL be x, LM be y and MP be z. x,y, and z are called the x, y, and z coordinates, of the point P in the space. In the fig. we can see that the point P (x, y, z) lies in the octant XOYZ and so all x, y, z are positive.

### Important Points to Remember

1. The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is given by

AB = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

2. The coordinates of the point R which divides the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) internally and externally in the ratio m : n are given by $\left ( \frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}, \frac{mz_{2}+nz_{1}}{m+n} \right )$

and $\left ( \frac{mx_{2}-nx_{1}}{m-n}, \frac{my_{2}-ny_{1}}{m-n}, \frac{mz_{2}-nz_{1}}{m-n} \right )$.

3. The mid-point of the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) are $\left ( \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2} \right )$.

4. The coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1) ,(x2, y2, z2) and (x3, y3, z3) are $\left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3} \right )$.

5. Equation of a line through P(x1, y1, z1) and having direction cosines l, m and n are

(x- x1)/ l = (y – y1)/ m = (z- z1)/ n.

6. Equation of a line through (x1, y1, z1) and (x2, y2, z2) is $\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}$.

## Solved Examples

Example 1.

Find the coordinates of the point which divides the line joining points (2, 3, 4) and (3, –4, 7) in ratio 3: 5 internally.

Solution:

Given points are (2, 3, 4) and (3, –4, 7).

m:n = 3:5

Let (x,y,z) be the required point, then

$\left ( \frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}, \frac{mz_{2}+nz_{1}}{m+n} \right )$

(mx2 + nx1)/(m + n) = (3×3 + 5×2)/(3+5) = (9+10)/8

= 19/8

(my2 + ny1)/(m + n) = (3×-4 + 5×3)/(3+5) = -12+15)/8 = ⅜

(mz2 + nz1)/(m + n) = (3×7 + 5×4)/(3+5) = (21+20)/8 = 41/8

Hence the required points are (19/8, ⅜, 41/8).

Example 2. Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, – 8) is divided by the YZ-plane.

Solution:

Consider YZ-plane divides the line segment joining A (4, 8, 10) and B (6, 10, -8) at P (x, y, z) in the ratio k : 1.

By using section formula, the coordinates of P are (6k + 4)/(k + 1), (10k + 8)/(k + 1), (-8k + 10)/(k + 1).

Given the point lies on YZ plane. So x coordinate is zero.

(6k + 4)/(k + 1) = 0

6k + 4 = 0

k = -4/6 = -2/3.

Hence the required ratio is 2:3.