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Calculation using vectors in 3D

In vector notation, the equations are as follows:

Equation for a sphere

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

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

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

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

• x {\displaystyle \mathbf {x} }  : points on the line
• o {\displaystyle \mathbf {o} }  : origin of the line
• d {\displaystyle d}  : distance from the origin of the line
• u {\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 d {\displaystyle d} , involving the dot product of vectors:

Equations combined ‖ o + d u − c ‖ 2 = r 2 ⇔ ( o + d u − c ) ⋅ ( o + d u − c ) = r 2 {\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: d 2 ( u ⋅ u ) + 2 d [ u ⋅ ( o − c ) ] + ( o − c ) ⋅ ( o − c ) − r 2 = 0 {\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) a d 2 + b d + c = 0 {\displaystyle ad^{2}+bd+c=0} where

• a = u ⋅ u = ‖ u ‖ 2 {\displaystyle a=\mathbf {u} \cdot \mathbf {u} =\left\Vert \mathbf {u} \right\Vert ^{2}}
• b = 2 [ u ⋅ ( o − c ) ] {\displaystyle b=2[\mathbf {u} \cdot (\mathbf {o} -\mathbf {c} )]}
• c = ( o − c ) ⋅ ( o − c ) − r 2 = ‖ o − c ‖ 2 − r 2 {\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 d = − 2 [ u ⋅ ( o − c ) ] ± ( 2 [ u ⋅ ( o − c ) ] ) 2 − 4 ‖ u ‖ 2 ( ‖ o − c ‖ 2 − r 2 ) 2 ‖ u ‖ 2 {\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 u {\displaystyle \mathbf {u} } is a unit vector, and thus ‖ u ‖ 2 = 1 {\displaystyle \left\Vert \mathbf {u} \right\Vert ^{2}=1} , we can simplify this further to (writing u ^ {\displaystyle {\hat {\mathbf {u} }}} instead of u {\displaystyle \mathbf {u} } to indicate a unit vector): ∇ = [ u ^ ⋅ ( o − c ) ] 2 − ( ‖ o − c ‖ 2 − r 2 ) {\displaystyle \nabla =[{\hat {\mathbf {u} }}\cdot (\mathbf {o} -\mathbf {c} )]^{2}-(\left\Vert \mathbf {o} -\mathbf {c} \right\Vert ^{2}-r^{2})} d = − [ u ^ ⋅ ( o − c ) ] ± ∇ {\displaystyle d=-[{\hat {\mathbf {u} }}\cdot (\mathbf {o} -\mathbf {c} )]\pm {\sqrt {\nabla }}}

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

See also

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

References

1. Eberly, David H. (2006). 3D game engine design: a practical approach to real-time computer graphics, 2nd edition. Morgan Kaufmann. p. 698. ISBN 0-12-229063-1. 0-12-229063-1 ↩

2. "Joachimsthal's Equation". http://mathworld.wolfram.com/JoachimsthalsEquation.html ↩