Game Engine Architecture

148 4. 3D Math for Games

The dot product can also be used to fi nd the height of a point above or

below a plane (which might be useful when writing a moon-landing game for

example). We can defi ne a plane with two vector quantities: a point Q lying

anywhere on the plane, and a unit vector n that is perpendicular (i.e., normal)

to the plane. To fi nd the height h of a point P above the plane, we fi rst calculate

a vector from any point on the plane (Q will do nicely) to the point in ques-

tion P. So we have v = P – Q. The dot product of vector v with the unit-length

normal vector n is just the projection of v onto the line defi ned by n. But that

is exactly the height we’re looking for. Therefore, h = v ⋅ n = (P – Q) ⋅ n. This

is illustrated in Figure 4.12.

4.2.4.8. Cross Product

The cross product (also known as the outer product or vector product) of two vec-

tors yields another vector that is perpendicular to the two vectors being multi-

plied, as shown in Figure 4.13. The cross product operation is only defi ned in

three dimensions:

Magnitude of the Cross Product

The magnitude of the cross product vector is the product of the magnitudes of

the two vectors and the sine of the angle between them. (This is similar to the

defi nition of the dot product, but it replaces the cosine with the sine.)

The magnitude of the cross product is equal to the area of the par-

allelogram whose sides are a and b, as shown in Figure 4.14. Since a triangle

is one-half of a parallelogram, the area of a triangle whose vertices are speci-

fi ed by the position vectors V 1 , V 2 , and V 3 can be calculated as one-half of the

magnitude of the cross product of any two of its sides:

a×b

a b

Figure 4.13. The cross
product of vectors a
and b (right-handed).

V 2

V 1

V 3

a= (V 2 – V 1 )

b= (V 3 – V 1 )

|a×b|

Figure 4.14. Area of a parallelogram expressed as the magnitude of a cross product.

[( ), ( ), ( )]

( ) ( ) ( ).

yz zy zx xz xy yx

yz zy zx xz xy yx

ab ab ab ab ab ab

ab ab ab ab ab ab

=−

=−

ab

i

− −

+−jk+−

×

triangle^12131

A = −×− 2 (VV VV) ( ).

a b ab×= sin( ) .θ

ab×