Graphics Reference
In-Depth Information
To prove (4), note that by definition
v
Ê
ˆ
1
Á
Á
Á
˜
˜
˜
M
vv
¥
¥◊◊◊¥
v
∑
w
=
det
(1.26)
1
2
n
-
1
v
w
n
Ë
¯
holds for
all
vectors
w
. When
w
is the vector
v
1
¥
v
2
¥ ···¥
v
n-1
we see that the left-
hand side of (1.26) is positive, which implies that the determinant is also. Now use
Lemma 1.6.4.
1.10.3. Proposition.
In
R
3
the generalized cross product agrees with the usual cross
product as defined in Section 1.5.
Proof.
This is Exercise 1.10.1.
The next proposition lists a few of the well-known properties of the cross product
in the special case of
R
3
.
The (generalized) cross product in
R
3
satisfies
1.10.4. Proposition.
(1) |
u
¥
v
| = |
v
| |
w
| sin q, where q is the angle between
u
and
v
.
(2)
u
¥ (
v
¥
w
) = (
u
•
w
)
v
- (
u
•
v
)
w
(
u
¥
v
) ¥
w
= (
u
•
w
)
v
- (
v
•
w
)
u
(3) |
u
¥
v
|
2
= |
u
|
2
|
v
|
2
- (
u
•
v
)
2
(4) (
u
1
¥
u
2
)•(
v
1
¥
v
2
) = (
u
1
•
v
1
)(
u
2
•
v
2
) - (
u
1
•
v
2
)(
u
2
•
v
1
)
Proof.
Exercise 1.10.2.
One way to look at identity (3) in Proposition 1.10.4 is that the cross product meas-
ures the deviation from equality in the Cauchy-Schwarz inequality.
1.10.5. Example.
Find the equation of the plane through (1,0,3) with basis
v
1
= (1,1,0) and
v
2
= (0,1,1).
Solution.
By Proposition 1.10.2(3),
(
)
uv v
=¥ =-
111
,,
1
2
is a normal vector for the plane. Therefore, an equation for it is
(
)
∑
(
(
)
-
(
)
)
=
111
,,
-
xyz
,,
103
,,
0
or
xyz
-+=4.
