Advanced Calculus Topics - Computer Graphics and Geometric Modeling - page 247

Proof. Apply Proposition 4.5.8 to the function g( p ) = f( p ) - c. Note that —g =—f.

We finish this section with two applications about the existence and uniqueness
of closest points between planes.

4.5.12. Theorem. Let X be a k-dimensional plane in R n . For any point p in R n there
is a unique point x in X that is closest to p . The point x is defined by the condition
that the vector px is orthogonal to X .

Proof. Let x 0 ΠX and assume that u 1 , u 2 ,..., u k are an orthonormal basis for X .
Parameterize the points of X with the function j : R k Æ X defined by

( ) =+

j tt , ,..., t xuu t + t ++ ... t u .

12 k 0 11 22 k k

Define differentiable functions s : R n Æ R and d : R k Æ R by

() =∑ s x p p

and

() = ( () ) =- () ( ) ∑- () ( )

d t sj t pt j pt j .

Clearly, finding a point x in X that is closest to p is equivalent to finding a minimum
of the function d.
Our first observation is that d does achieve a minimum and that this minimum is
a critical point of d. This follows from Theorems 4.5.1 and 4.5.2. The domain of d,
R k , is of course not a compact set, but we can apply the theorems to the function d
restricted to some large closed disk in R k with the property that d is larger at every
point on the boundary of that disk than at some point on its interior. This will guar-
antee that a relative extremum will occur in the interior of the disk. Such a disk clearly
exists because d( t ) goes to infinity as | t | goes to infinity.
Next, we show that d has a unique minimum, one defined by the stated orthogo-
nality condition. The chain rule applied to d gives that

∂
∂

d
t i

() =- ( - () ) t 2 upt ∑ j .

i

Since the critical points t of d are the points where all of these partials vanish, we see
that solving for those points is equivalent to solving for those points x =j( t ) in X sat-
isfying u i • xp = 0, for all i, which shows that xp must be orthogonal to X .
Finally, we need to show that a point of X defined by such orthogonality condi-
tions is unique. See Figure 4.16. Let x ¢=j( t ¢) and assume that u i • x ¢ p = 0, for all i.
This would imply that u i • xx ¢=0, for all i. Since xx ¢ is a vector in the plane X and the
u i form a basis for X , we must have that xx ¢= 0 , that is, x = x ¢. The theorem is proved.

Theorem 4.5.12 is a special case of the next theorem. We sketched the proof in
this special case because, being simpler, it brought out more clearly the essential steps
in the proof of these types of theorems.