Linear Algebra Topics - Computer Graphics and Geometric Modeling - page 4

-= +
+=-
-+ =-+

1 2
212
1

t s

t s

t 2 s

for s and t. The first two equations imply that t =-1 and s = 0. Since these two values
also satisfy the third equation, the lines L 1 and L 2 intersect at the point (2,1,-2).

Note. A common mistake when trying to solve a problem like that in Example 1.2.2
is to use the same variable for both s and t. Just because lines intersect does not mean
that persons “walking” along the lines will get to the intersection point at the same
“time.”

Definition. Points are said to be collinear if they lie on the same line and non-
collinear , otherwise.

Let p , q ΠR n . The set Definition.

{

Π[]

}

ppq + t t , (1.7)

is called the segment from p to q and is denoted by [ p , q ]. The points of [ p , q ] are said
to lie between p and q .

Note that [ p , q ] = [ q , p ] (Exercise 1.2.5). A segment basically generalizes the notion
of a closed interval of the real line, which explains the notation, but the two concepts
are not quite the same when n = 1 (Exercise 1.2.6). The following proposition gives a
very useful alternative characterization of a segment.

Let p , q ΠR n . Then

1.2.3. Proposition.

{ }

[ ]

n

p q , = x ΠR px + xq = pq . (1.8)

Proof. Let

{ } .

S = x px + xq = pq

In order to show that [ p , q ] = S we must prove the two inclusions [ p , q ] Õ S and
S Õ [ p , q ].
To prove that [ p , q ] Õ S , let x Œ [ p , q ]. Then x = p + t pq for some t with 0 £ t £ 1.
It follows that

px += xq t pq +- 1 t pq = pq ,

so that x ΠS .
To prove that S Õ [ p , q ], let x Œ S . Since | px | + | xq | = | pq | = | px + xq |, the triangle
inequality implies that the vectors px and xq are linearly dependent. Assume without
loss of generality that px = t xq . Then