Line–sphere intersection

The three possible line-sphere intersections:
1. No intersection.
2. Point intersection.
3. Two point 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 $d$:

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 .)

$a d^2 + b d + c = 0$

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

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

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).

See also

• Analytic geometry
• Line-plane intersection
• Line of intersection between two planes

References

• David H. Eberly (2006), 3D game engine design: a practical approach to real-time computer graphics, 2nd edition, Morgan Kaufmann. ISBN 0-12-229063-1