Unlike two-dimensional geometry which deals with flat shapes like lines, circles, and polygons, three-dimensional geometry deals with solid shapes like cubes, spheres, and cylinders.

The main components to consider in 3D space are the x-axis, y-axis, and z-axis. These three axes are perpendicular to each other

What is Cartesian Three-dimensional Geometry

We require three mutually perpendicular axes to locate a point in the space.

The space is called Cartesian space and axis are called the coordinates axis

The axes are called the x, y and z-axes

The point of intersection of axis is called origin.

The three planes determined by the pair of axes are the coordinate planes, called XY, YZ and ZX-planes

The three coordinate planes divide the space into eight parts known as octants

The coordinates of a point P in three-dimensional geometry is always written in the form of triplet like (x, y, z). Here x, y and z are the distances from the YZ, ZX and XY-planes

A point on the x –axis is of the form (x,0,0)

A point on the y –axis is of the form (0, y,0)

A point on the y –axis is of the form (0,0, z)

Sign of the coordinates

The coordinates of the points in the octants will have sign per the below table

Distance formula

Let P and Q be the point in the 3D space

Through the points P and Q draw planes parallel to the coordinate planes so as to form a rectangular parallelopiped with one diagonal PQ

In $\Delta PAQ$, $\angle PAQ=90^0$,So
$PQ^2= PA^2+ AQ^2$ --(1)

In triangle ANQ is right angle triangle with $\angle ANQ$ a right angle
$AQ^2= AN^2+ NQ^2$ -(2)

From 1 and 2
$PQ^2= PA^2+ AN^2+ NQ^2$

Now $PA = y_2- y_1$, $AN = x_2- x_1$ and $NQ = z_2 – z_1$

Distance of Point from Origin (0,0,0)
$D= \sqrt {x^2 + y^2 + x^2}$

Section Formula

A point P (x, y, z) which divide the line segment AB in the ratio m and n internally is given by
$x=\frac {(mx_2+nx_1)}{(m+n)}$
$y=\frac {(my_2+ny_1)}{(m+n)}$
$z=\frac {(mz_2+nz_1)}{(m+n)}$

The midpoint P is given by
$((x_1+x_2)/2)$,$((y_1+y_2)/2)$,$((z_1+z_2)/2)$
A point P (x, y, z) which divide the line segment AB in the ratio m and n externally is given by
$x=\frac {(mx_2-nx_1)}{(m-n)}$
$y=\frac {(my_2-ny_1)}{(m-n)}$
$z=\frac {(mz_2-nz_1)}{(m-n)}$
We can use the above in the ratio k: 1 are obtained
by taking k=m/n
$x=\frac {(kx_2+x_1)}{(k+1)}$
$y=\frac {(ky_2+y_1)}{(k+1)}$
$z=\frac {(kz_2+z_1)}{(k+1)}$

Centroid of Triangle:

Centroid of triangle ABC of coordinates $A(x_1,y_1, z_1)$ , $B(x_2,y_2,z_2)$ and $C(x_3,y_3,z_3)$
$((x_1+x_2+x_3)/3)$,$((y_1+y_2+y_3)/3)$,$((z_1+z_2+z_3)/3)$