Line–sphere intersection

In vector notation, the equations are as follows:

Equation for a sphere

${\displaystyle \left\Vert \mathbf {x} -\mathbf {c} \right\Vert ^{2}=r^{2}}$

• ${\displaystyle \mathbf {c} }$ - center point
• ${\displaystyle r}$ - radius
• ${\displaystyle \mathbf {x} }$ - points on the sphere

Equation for a line starting at ${\displaystyle \mathbf {o} }$

${\displaystyle \mathbf {x} =\mathbf {o} +d\mathbf {u} }$

• ${\displaystyle d}$ - distance along line from starting point
• ${\displaystyle \mathbf {u} }$ - direction of line (a unit vector)
• ${\displaystyle \mathbf {o} }$ - origin of the line
• ${\displaystyle \mathbf {x} }$ - points on the line

Searching for points that are on the line and on the sphere means combining the equations and solving for ${\displaystyle d}$, involving the dot product of vectors:

Equations combined

${\displaystyle \left\Vert \mathbf {o} +d\mathbf {u} -\mathbf {c} \right\Vert ^{2}=r^{2}\Leftrightarrow (\mathbf {o} +d\mathbf {u} -\mathbf {c} )\cdot (\mathbf {o} +d\mathbf {u} -\mathbf {c} )=r^{2}}$

Expanded

${\displaystyle d^{2}(\mathbf {u} \cdot \mathbf {u} )+2d(\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} ))+(\mathbf {o} -\mathbf {c} )\cdot (\mathbf {o} -\mathbf {c} )=r^{2}}$

Rearranged

${\displaystyle d^{2}(\mathbf {u} \cdot \mathbf {u} )+2d(\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} ))+(\mathbf {o} -\mathbf {c} )\cdot (\mathbf {o} -\mathbf {c} )-r^{2}=0}$

The form of a quadratic formula is now observable. (This quadratic equation is an instance of Joachimsthal's equation.[2])

${\displaystyle ad^{2}+bd+c=0}$

where

• ${\displaystyle a=\mathbf {u} \cdot \mathbf {u} =\left\Vert \mathbf {u} \right\Vert ^{2}}$
• ${\displaystyle b=2(\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} ))}$
• ${\displaystyle c=(\mathbf {o} -\mathbf {c} )\cdot (\mathbf {o} -\mathbf {c} )-r^{2}=\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2}}$

Simplified

${\displaystyle d={\frac {-2(\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} ))\pm {\sqrt {(2(\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} )))^{2}-4\left\Vert \mathbf {u} \right\Vert ^{2}(\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2})}}}{2\left\Vert \mathbf {u} \right\Vert ^{2}}}}$

Note that ${\displaystyle \mathbf {u} }$ is a unit vector, and thus ${\displaystyle \left\Vert \mathbf {u} \right\Vert ^{2}=1}$. Thus, we can simplify this further to

${\displaystyle d=-(\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} ))\pm {\sqrt {\nabla }}}$

${\displaystyle \nabla =(\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} ))^{2}-(\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2})}$

• If ${\displaystyle \nabla <0}$, then it is clear that no solutions exist, i.e. the line does not intersect the sphere (case 1).
• If ${\displaystyle \nabla =0}$, then exactly one solution exists, i.e. the line just touches the sphere in one point (case 2).
• If ${\displaystyle \nabla >0}$, two solutions exist, and thus the line touches the sphere in two points (case 3).

• Analytic geometry
• Line–plane intersection
• Plane–plane intersection