Three Dimensional (3-D) Coordinate System

A single point in a plane needs only an ordered pair $(a,b)$ (its signed distances from the Y- and X-axes).

Locating a point in space, however, needs three numbers: three mutually perpendicular coordinate axes

$OX,OY,OZ$ through a common origin $O$ are set up, with the positive direction of the Z-axis fixed by the

right-hand rule -- curl the fingers of the right hand from the positive X-axis towards the positive

Y-axis, and the thumb then points along the positive Z-axis. A point in space is written as an ordered triple

$(x,y,z)$.

For a point $P(x,y,z)$: dropping perpendiculars from $P$ to the three coordinate planes lands at feet

$L$ (in the XY-plane), $M$ (in the YZ-plane), $N$ (in the XZ-plane); the origin is $O(0,0,0)$; points on the

coordinate axes look like $A(x,0,0),B(0,y,0),C(0,0,z)$; and points on the coordinate planes look like

$L(x,y,0),M(0,y,z),N(x,0,z)$.

Distances. The distance of $P(x,y,z)$ from the XY-plane is $|z|$, from the YZ-plane is $|x|$, from the

XZ-plane is $|y|$. Repeated use of the Pythagorean theorem (first in right triangle $OLP$, then in right

triangle $OAL$) gives the distance of $P$ from the origin as $|OP|=\sqrt{x^2+y^2+z^2}$, and the distance

between two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ as

$|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. The distance of $P(x,y,z)$ from the X-axis works out to

$\sqrt{y^2+z^2}$ by the same method (and similarly $\sqrt{x^2+z^2}$ from the Y-axis, $\sqrt{x^2+y^2}$ from the

Z-axis).

Octants. The three coordinate planes cut space into eight regions called octants, each with its own fixed

sign pattern for $(x,y,z)$: Octant I $(+,+,+)$, II $(-,+,+)$, III $(+,-,+)$, IV $(-,-,+)$, V $(+,+,-)$, VI

$(-,+,-)$, VII $(+,-,-)$, VIII $(-,-,-)$ -- octants II, III, IV are reached by rotating anticlockwise around

the positive Z-axis, and V is directly below I, with VI, VII, VIII obtained the same way around the negative

Z-axis. …