# Co-ordinate of a Point in Three Dimension

Three-dimensional space that can also be known as 3-space or tri-dimensional space.

It is a geometric setting which contains three values are required to determine the position of an element .In physics and mathematics, a sequence of n numbers can be understood as a location in n-dimensional space. When n = 3 it is called three-dimensional Euclidean space. It is commonly represented by the symbol ℝ3. This acts as a three-parameter model of the physical universe in which all known matter exists. This space is only one example of a large spaces in three dimensions called 3-manifolds. In this case, these three values are chosen from the terms width, height, depth, and length.

Points in 3 Dimension

On a two dimensional plane a point in the xy-plane by an ordered pair that consists of two real numbers, an x-coordinate and y-coordinate, which denote signed distances along the x-axis and y-axis, respectively, from the origin, which is the point (0, 0). These axes, which are referred to as the coordinate axes, divided the plane into four quadrants. The properties of three-dimensional space.

• a point is represented by an ordered triple (x, y, and z) that consists of three numbers, an x-coordinate, a y-coordinate,

•  A z-coordinate in the two-dimensional xy-plane, these coordinates indicate the signed distance along the coordinate axes,

•  The x-axis, y-axis and z-axis, respectively, from the origin, denoted by o, which has coordinates (0, 0, and 0).

There is a one-to-one correspondence between a point in xyz-space and a triple in R3, which is the set of all ordered triples of real numbers. This is known as a three-dimensional rectangular coordinate system.

Example

The figure displays the point (2, 3, and 1) in xyz-space, denoted by the letter P, along with its projections onto the coordinate planes .The origin is denoted by the letter o.

The point (2, 3, 1) in xyz-space, denoted by the letter P. The origin is denoted by the letter o. The projections of P onto the coordinate planes are indicated by the diamonds. The dashed lines are line segments perpendicular to the coordinate planes that connect P to its projections. Just as the x-axis and y-axis divide the xy-plane into four quadrants, these three planes divide xyz-space into eight octants. Within each octant, all x-coordinates have the same sign, as do all y-coordinates, and all z-coordinates

How to Find Coordinates of a Point in a Three Dimensional Space

Finding a point in x,y,z-space can be difficult because, unlike graphing in the x,y-plane, depth perception is required. The projection of a point (x, y, z) onto the x,y-plane is obtained by connecting the point to the x,y-plane by a line segment that is perpendicular to the plane, and computing the intersection of the line segment with the plane. Similarly, the projection of this point onto the xy-plane is the point (0, y, z), and the projection of this point onto the xz-plane is the point (x, 0, z). The figure shows these projections, and how they can be used to plot a point in x,y,z-space. One can first plot the point’s projections, which is similar to the task of plotting points in the x,y-plane, and then use line segments originating from these projections and perpendicular to the coordinate planes to “locate” the point in x,y,z-space.

The Distance Formula Between the Two Points in Three Dimension

The distance between two points P1 = (x1, y1) and P2 = (x2, y2) in the xy-plane is given by the distance formula,

d (P1, P2) = $\sqrt{(x2 − x1)^{2} + (y2 − y1)^{2}}$

Similarly, the distance between two points P1 = (x1, y1, z1) and P2 = (x2, y2, z2) in xyz-space is given by the following generalization of the distance formula,

d (P1, P2) = $\sqrt{(x2 − x1)^{2} + (y2 − y1)^{2} + (z2 − z1)^{2}}$

This can be proved with the application of Pythagorean Theorem.

Solved Examples –

Question: Find the distance between P1 = (2, 3, 1) and P2 = (8, −5, 0)

Solution:

From the distance formula, we have.

d (P1, P2) =$\sqrt{(8 − 2)^{2} + (-5 − 3)^{2} + (0 − 1)^{2}}$

= $\sqrt{36 + 64 + 1}$

= $\sqrt{101}$ ≈ 10.05.

Question: Find the distance between the points (2,-5, and 7) and (3, 4, 5).

Solution: d ​= $\sqrt{(3 − 2)^{2} + (4-(-5))^{2} + (5 − 7)^{2}}$

=  $\sqrt{1+81+4}$

=   $\sqrt{86}$