MATH 152 Chapter 11.1

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3D Coordinates

2D space was the set of all points $\left(x_{0},\,y_{0}\right)$, where $x_{0}$ refers to the horizontal and $y_{0}$ refers to the vertical.

Now 3D space is the set of all points $\left(x_{0},\,y_{0},\,z_{0}\right)$, where $z_{0}$ is height, $y_{0}$ is width, and $x_{0}$ is depth.

Distance between 2 points in 3D space

$d={\sqrt {\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}+\left(\Delta z\right)^{2}}}$

Other 3D Terms

The set of points equidistant from a given point is now called a sphere

all points $\left(x,\,y,\,z\right)$ such that their distance to the center $\left(x_{0},\,y_{0},\,z_{0}\right)$ is radius $r$:

$\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}+\left(z-z_{0}\right)^{2}=r^{2}$

Midpoint in 3D space

$\left({\frac {x_{1}+x_{2}}{2}},\,{\frac {z_{1}+z_{2}}{2}},\,{\frac {z_{1}+z_{2}}{2}}\right)$

Example 1

Describe the set of all points which are equidistant from the TWO points (1, 2, 3) and (3, 5, 7). Find the equation of this object.

${\begin{aligned}{\sqrt {\left(x-1\right)^{2}+\left(y-2\right)^{2}+\left(z-3\right)^{2}}}=&{\sqrt {\left(x-3\right)^{2}+\left(y-5\right)^{2}+\left(z-7\right)^{2}}}\\4x+6y+8z=&69\end{aligned}}$

Similar to the perpendicular bisector in 2D.

Example 2

Is $A\left(3,\,-4,\,1\right)$, $B\left(5,\,-3,\,0\right)$, $C\left(6,\,-7,\,4\right)$ a right triangle?

${\begin{aligned}AB^{2}&=\left(5-3\right)^{2}+\left(-3+4\right)^{2}+\left(0-1\right)^{2}=6\\BC^{2}&=\left(6-5\right)^{2}+\left(-7+3\right)^{2}+\left(4-0\right)^{2}=33\\AC^{2}&=\left(6-3\right)^{2}+\left(-7+4\right)^{2}+\left(4-1\right)^{2}=27\\BC^{2}&{\overset {\underset {\mathrm {?} }{}}{=}}AB^{2}+AC^{2}\quad (true)\end{aligned}}$

Yes, these points do describe a right triangle.