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}

See also

References

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