Spheres and lines

There are three possible relative positions for a sphere and a line in space.

A sphere and a line can therefore have one, two or no common point.

The individual coordinates are used in the equation of a sphere to calculate the intersection points.

Example

$g: \vec{x} = \begin{pmatrix} 5 \\ 6 \\ 5 \end{pmatrix} + r \cdot \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$

$k: (x+1)^2+(y-2)^2$ $+(z-1)^2=17$

1. Break $g$ into 3 equations

   We replace $\vec{x}$ and write out the respective coordinates as our own equation.

   $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 5 \\ 6 \\ 5 \end{pmatrix} + r \cdot \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$

   1. $x=5+3r$
   2. $y=6+2r$
   3. $z=5+2r$

2. Insert coordinates

   The equations are now used in the equation of a sphere for $x$, $y$ and $z$.

   $(x+1)^2+(y-2)^2$ $+(z-1)^2=17$

   $(5+3r+1)^2$ $+(6+2r-2)^2$ $+(5+2r-1)^2=17$

   $(6+3r)^2$ $+(4+2r)^2$ $+(4+2r)^2=17$

   Use binomial theorem to resolve parentheses

   $36+36r+9r^2$ $+16+16r+4r^2$ $+16+16r+4r^2=17$

   $17r^2+68r+68=17\quad|-17$

   $17r^2+68r+51=0\quad|:17$

   $r^2+4r+3=0$

   Use quadratic formula I to solve quadratic equation.

   $r_{1,2}=-\frac{p}2\pm\sqrt{(\frac{p}2)^2-q}$
   $r_{1,2}=-2\pm\sqrt{2^2-3}$
   $r_{1,2}=-2\pm1$

   $r_{1}=-1$ and $r_{2}=-3$

3. Insert $r$

   The two calculated $r$ are inserted into the equation of a line in order to obtain the intersection points.

   $\vec{OS_1} = \begin{pmatrix} 5 \\ 6 \\ 5 \end{pmatrix} - 1 \cdot \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$ $=\begin{pmatrix} 2 \\ 4 \\ 3 \end{pmatrix}$

   $\vec{OS_2} = \begin{pmatrix} 5 \\ 6 \\ 5 \end{pmatrix} - 3 \cdot \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$ $=\begin{pmatrix} -4 \\ 0 \\ -1 \end{pmatrix}$

   It is a secant that intersects the sphere at $S_1(2|4|3)$ and $S_2(-4|0|-1)$.