Graphics Reference
In-Depth Information
We can now define the direct path
−
AC in terms of two indirect paths:
−
AC
=
−
AB
+
−
BC
and
−
AC
=
−
AD
+
−
DC
therefore,
−
AB
+
−
BC
=
−
AD
+
−
DC
Or, using vector names,
r
+
v
s
=
s
+
u
r
Because these vectors are not collinear, we can equate coefficients and state that
=
=
1
u and v
1
=
−
BC, which means that their lengths are equal. Hence
we have proved that the opposite sides of a parallelogram are equal.
Let's try another example and prove that the diagonals of a parallelogram bisect each other.
1,
−
AB
=
−
DC and
−
AD
If u
=
1 and v
=
r
D
C
E
s
s
A
B
r
Figure 1.13.
=
−
AB
=
−
DC
Figure 1.13 shows a parallelogram ABCD formed from the vectors
r
and
s
, where
r
=
−
AD
=
−
BC. We can see that
and
s
−
AC
=
r
+
s
(1.2)
and
−
DB
s
(1.3)
We observe that the diagonals intersect at E, which means that
−
AE and
−
AC are collinear.
Therefore, we can state
=
r
−
−
AE
−
AC
=
(1.4)
where is some scalar.
