Line–sphere intersection

In vector notation, the equations are as follows:

Equation for a sphere

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

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

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

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

• ${\displaystyle \mathbf {x} }$ : points on the line
• ${\displaystyle \mathbf {o} }$ : origin of the line
• ${\displaystyle d}$ : distance from the origin of the line
• ${\displaystyle \mathbf {u} }$ : direction of line (a non-zero vector)

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 and 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 in the specific case where ${\displaystyle \mathbf {u} }$ is a unit vector, and thus ${\displaystyle \left\Vert \mathbf {u} \right\Vert ^{2}=1}$, we can simplify this further to (writing ${\displaystyle {\hat {\mathbf {u} }}}$ instead of ${\displaystyle \mathbf {u} }$ to indicate a unit vector):

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

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

• 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).

• Intersection (geometry) § A line and a circle
• Analytic geometry
• Line–plane intersection
• Plane–plane intersection
• Plane–sphere intersection