# Line–sphere intersection

In analytic geometry, a line and a sphere can intersect in three ways: no intersection at all, at exactly one point, or in two points. Methods for distinguishing these cases, and determining equations for the points in the latter cases, are useful in a number of circumstances. For example, this is a common calculation to perform during ray tracing (Eberly 2006:698).

## Calculation using vectors in 3D

In vector notation, the equations are as follows:

Equation for a sphere

$\left\Vert \mathbf {x} -\mathbf {c} \right\Vert ^{2}=r^{2}$ • $\mathbf {c}$ - center point
• $r$ - radius
• $\mathbf {x}$ - points on the sphere

Equation for a line starting at $\mathbf {o}$ $\mathbf {x} =\mathbf {o} +d\mathbf {l}$ • $d$ - distance along line from starting point
• $\mathbf {l}$ - direction of line (a unit vector)
• $\mathbf {o}$ - origin of the line
• $\mathbf {x}$ - points on the line

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

Equations combined
$\left\Vert \mathbf {o} +d\mathbf {l} -\mathbf {c} \right\Vert ^{2}=r^{2}\Leftrightarrow (\mathbf {o} +d\mathbf {l} -\mathbf {c} )\cdot (\mathbf {o} +d\mathbf {l} -\mathbf {c} )=r^{2}$ Expanded
$d^{2}(\mathbf {l} \cdot \mathbf {l} )+2d(\mathbf {l} \cdot (\mathbf {o} -\mathbf {c} ))+(\mathbf {o} -\mathbf {c} )\cdot (\mathbf {o} -\mathbf {c} )=r^{2}$ Rearranged
$d^{2}(\mathbf {l} \cdot \mathbf {l} )+2d(\mathbf {l} \cdot (\mathbf {o} -\mathbf {c} ))+(\mathbf {o} -\mathbf {c} )\cdot (\mathbf {o} -\mathbf {c} )-r^{2}=0$ The form of a quadratic formula is now observable. (This quadratic equation is an example of Joachimsthal's Equation .)
$ad^{2}+bd+c=0$ where
• $a=\mathbf {l} \cdot \mathbf {l} =\left\Vert \mathbf {l} \right\Vert ^{2}$ • $b=2(\mathbf {l} \cdot (\mathbf {o} -\mathbf {c} ))$ • $c=(\mathbf {o} -\mathbf {c} )\cdot (\mathbf {o} -\mathbf {c} )-r^{2}=\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2}$ Simplified
$d={\frac {-2(\mathbf {l} \cdot (\mathbf {o} -\mathbf {c} ))\pm {\sqrt {(2(\mathbf {l} \cdot (\mathbf {o} -\mathbf {c} )))^{2}-4\left\Vert \mathbf {l} \right\Vert ^{2}(\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2})}}}{2\left\Vert \mathbf {l} \right\Vert ^{2}}}$ Note that $\mathbf {l}$ is a unit vector, and thus $\left\Vert \mathbf {l} \right\Vert ^{2}=1$ . Thus, we can simplify this further to
$d=-(\mathbf {l} \cdot (\mathbf {o} -\mathbf {c} ))\pm {\sqrt {(\mathbf {l} \cdot (\mathbf {o} -\mathbf {c} ))^{2}-(\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2})}}$ • If the value under the square-root ($(\mathbf {l} \cdot (\mathbf {o} -\mathbf {c} ))^{2}-(\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2})$ ) is less than zero, then it is clear that no solutions exist, i.e. the line does not intersect the sphere (case 1).
• If it is zero, then exactly one solution exists, i.e. the line just touches the sphere in one point (case 2).
• If it is greater than zero, two solutions exist, and thus the line touches the sphere in two points (case 3).