# Three-Dimensional Coordinate System

• Points in space: $$A,$$ $$B,$$ $$C,$$ $$D,$$ $${P_1},$$ $${P_2},$$ $${P_3},$$ $${P_4}$$
Point coordinates: $$\left( {{x_0},{y_0},{z_0}} \right),$$ $$\left( {{x_1},{y_1},{z_1}} \right),$$ $$\left( {{x_2},{y_2},{z_2}} \right),$$ $$\left( {{x_3},{y_3},{z_3}} \right),$$ $$\left( {{x_4},{y_4},{z_4}} \right)$$
Real number: $$\lambda$$
Distance between two points: $$d$$
Area of a triangle: $$S$$
Volume of a pyramid: $$V$$
1. A three-dimensional Cartesian coordinate system is formed by a point called the origin (denoted by $$O$$) and a basis consisting of three mutually perpendicular vectors. These vectors define the three coordinate axes: the $$x-,$$ $$y-,$$ and $$z-$$axis. They are also known as the abscissa, ordinate and applicate axis, respectively. The coordinates of any point in space are determined by three real numbers: $$x$$, $$y$$, $$z$$.
2. The distance between two points $$A\left( {{x_1},{y_1},{z_1}} \right)$$ and $$B\left( {{x_2},{y_2},{z_2}} \right)$$ in space is determined by the formula
$$d = \left| {AB} \right| =$$ $$\Big[ {{{\left( {{x_2} – {x_1}} \right)}^2} }$$ $$+\;{ {{\left( {{y_2} – {y_1}} \right)}^2} }$$ $$+\;{ {{\left( {{z_2} – {z_1}} \right)}^2}}\Big]^{\large{\frac{1}{2}}\normalsize}$$
3. Dividing a line segment in the ratio $$\lambda$$
Suppose that the point $$C\left( {{x_0},{y_0},{z_0}} \right)$$ divides the segment $$AB$$ in the ratio $$\lambda$$. The coordinates of the point $$C$$ are given by the expressions
$${x_0} = {\large\frac{{{x_1} + \lambda {x_2}}}{{1 + \lambda }}\normalsize},\;$$ $${y_0} = {\large\frac{{{y_1} + \lambda {y_2}}}{{1 + \lambda }}\normalsize},\;$$ $${z_0} = {\large\frac{{{z_1} + \lambda {z_2}}}{{1 + \lambda }}\normalsize},$$ where $$\lambda = {\large\frac{{AC}}{{CB}}\normalsize},\;$$ $$\lambda \ne – 1,$$
where $${x_1}$$, $${y_1}$$, $${z_1}$$ are the coordinates of the point $$A$$, and $${x_2}$$, $${y_2}$$, $${z_2}$$ are the coordinates of the point $$B$$.
4. The coordinates of the midpoint of the line segment are obtained from the previous formulas at $$\lambda = 1$$ and are written as
$${x_0} = {\large\frac{{{x_1} + {x_2}}}{2}\normalsize},\;$$ $${y_0} = {\large\frac{{{y_1} + {y_2}}}{2}\normalsize},\;$$ $${z_0} = {\large\frac{{{z_1} + {z_2}}}{2}\normalsize},$$, where $$\lambda = {\large\frac{{AC}}{{CB}}\normalsize} = 1.$$
5. Area of a triangle
The area of a triangle with the vertices $${P_1}\left( {{x_1},{y_1},{z_1}} \right),$$ $${P_2}\left( {{x_2},{y_2},{z_2}} \right),$$ $${P_3}\left( {{x_3},{y_3},{z_3}} \right)$$ is found by the formula
$$S =$$ $${\large\frac{1}{2}\normalsize}\left[ {{{\left| {\begin{array}{*{20}{c}} {{y_1}} & {{z_1}} & 1\\ {{y_2}} & {{z_2}} & 1\\ {{y_3}} & {{z_3}} & 1 \end{array}} \right|}^2} }\right.+\left.{ {{\left| {\begin{array}{*{20}{c}} {{z_1}} & {{x_1}} & 1\\ {{z_2}} & {{x_2}} & 1\\ {{z_3}} & {{x_3}} & 1 \end{array}} \right|}^2} }\right.+\left.{ {{\left| {\begin{array}{*{20}{c}} {{x_1}} & {{y_1}} & 1\\ {{x_2}} & {{y_2}} & 1\\ {{x_3}} & {{y_3}} & 1 \end{array}} \right|}^2}}\right] .$$
6. Volume of a pyramid
The volume of a pyramid whose vertices have the coordinates $${P_1}\left( {{x_1},{y_1},{z_1}} \right)$$, $${P_2}\left( {{x_2},{y_2},{z_2}} \right)$$, $${P_3}\left( {{x_3},{y_3},{z_3}} \right)$$, $${P_4}\left( {{x_4},{y_4},{z_4}} \right)$$ is determined by the expression
$$V =$$ $$\pm {\large\frac{1}{6}\normalsize}\left| {\begin{array}{*{20}{c}} {{x_1}} & {{y_1}} & {{z_1}} & 1\\ {{x_2}} & {{y_2}} & {{z_2}} & 1\\ {{x_3}} & {{y_3}} & {{z_3}} & 1\\ {{x_4}} & {{y_4}} & {{z_4}} & 1 \end{array}} \right|$$ or $$V =$$ $$\pm {\large\frac{1}{6}\normalsize}\left| {\begin{array}{*{20}{c}} {{x_1} – {x_4}} & {{y_1} – {y_4}} & {{z_1} – {z_4}}\\ {{x_2} – {x_4}} & {{y_2} – {y_4}} & {{z_2} – {z_4}}\\ {{x_3} – {x_4}} & {{y_3} – {y_4}} & {{z_3} – {z_4}} \end{array}} \right|.$$

The sign in the right side of the formulas above is chosen so that to get a positive value for the area or volume.