Picking - Introduction to 3D Game Programming with DirectX 9.0 - page 263

To determine if and where a ray p ( t )= p 0 + t u intersects a sphere, we
plug the ray into an implicit sphere equation and solve for the parame-
ter t that satisfies the sphere equation, giving us the parameter that
yields the intersection point(s).
Plugging the ray into the sphere equation:

p t c r 0

p t u c r 0

0

...from which we obtain the quadratic equation:

2

At Bt C 0

where A = u · u , B =2( u ·( p 0 - c )), and C =( p 0 - c )·( p 0 - c )- r 2 .If u
is normalized, then A =1.
Assuming u is normalized, we solve for t 0 and t 1 :

B B

2

4 C B B

2

4 C

t t

0 1

2

2

Figure 15.4 shows the possible results for t 0 and t 1 and shows what
these results mean geometrically.

Figure 15.4: a) The ray misses the sphere; both t 0 and t 1 will result in
imaginary solutions. b) The ray is in front of the sphere; both t 0 and t 1
will be negative. c) The ray is inside the sphere; one of the solutions
will be positive and one will be negative. The positive solution yields
the single intersection point. d) The ray intersects the sphere; both t 0
and t 1 are positive. e) The ray is tangent to the sphere, in which case
the solutions are positive and t 0 = t 1 .

The following method returns true if the ray passed in intersects the
sphere passed in. It returns false if the ray misses the sphere:
bool PickApp::raySphereIntersectionTest(Ray* ray,
BoundingSphere* sphere)

{

D3DXVECTOR3 v = ray->_origin - sphere->_center;