Transformations - Transformations and Projections in Computer Graphics - page 38

If we omit the absolute value, then the distance becomes a signed quantity. We can
think of the plane as if it divides all of space into two parts, one in the direction of N
and the other on the other side of the plane. The distance is positive if P is located in
that part of space pointed to by the normal (which is the case in Figure 1.21), and it is
negative in the opposite case.

Exercise 1.40: What's the distance of a plane from the origin?

Now that we can figure out the distance between a point and a plane, the last
step is to reflect a point about a given plane. We start with a point P =( x, y, z )

and a plan e Ax + By + C z + D = 0. We denote the normal unit vector by N =

( A, B, C ) / √ A 2 + B 2 + C 2 and the (signed) distance between P and the plane by d .To
get from P totheplane,wehavetotraveladistance d in the direction of N . To arrive
at the reflection point P ∗ , we should travel another d units in the same direction. Thus,
the reflection P ∗ of P is given by

2( Ax + By + Cz + D )
A 2 + B 2 + C 2

P ∗ = P

− 2 d N = P − ( A, B, C ) . (1.28)

Exercise 1.41: Why P − 2 d N and not P +2 d N ?

Most neurotics have been mindful of their five W's since grammar school: why, why,
why, why, why.

—Terri Guillemets

y y

(1,1,1)

(0,1,2)

(1,1,1)

z

( − 1,0,2)

x x

( − 1, − 1,1)

(a)

(b)

Figure 1.22: Reflection in Three Dimensions: Examples.

Examples . We select (Figure 1.22a) the plane x + y =0andthepoint P =(1 , 1 , 1).
Equation (1.28) becomes

2(1 + 1)
1+1+0 (1 , 1 , 0) = (

P ∗ =(1 , 1 , 1)

− − 1 , − 1 , 1) .

Similarly, point P =(0 , 1 , 2) is reflected to

2(0 + 1)
1+1+0 (1 , 1 , 0) = (

P ∗ =(0 , 1 , 2) − − 1 , 0 , 2) .