To determine the direction of
u
×
v
we have the right hand rule; Fingers point the
first vector, Palm points the second vector, tumb gives the direction of the cross product.
Theorem 3.5.1 (Relationship involving Cross Product and Dot Prod
uct:)
If u,v and w are vectors in 3space, then
(a)
u
·
(
u
×
v
) = 0
(that is
u
×
v
and u are orthogonal)
(b)
v
·
(
u
×
v
) = 0
(that is
u
×
v
and v are orthogonal)
(c)

u
×
v

2
=

u

2

v

2

(
u
·
v
)
2
(Lagrange’s Identity)
(d)
u
×
(
v
×
w
) = (
u
·
w
)
v

(
u
·
v
)
w
(e)
(
u
×
v
)
×
w
= (
u
·
w
)
v

(
v
·
w
)
u
Proof:
Write
u
= (
u
1
, u
2
, u
3
)
,
v
= (
v
1
, v
2
, v
3
)
and apply the definitions of dot prod
uct and cross product.
Importance:
•
From part (a) and (b) we see that if one needs to find a vector that is orthogonal
to both vectors u and v then
u
×
v
will do the work.
•
We will see that this property can be used to write a plane equation.
Example:
Let
u
= (1
,
0
,
2)
, v
= (

1
,
1
,
3)
then
u
×
v
=
i
j
k
1
0
2

1
1
3
=

2
i

5
j
+
k
=
(

2
,

5
,
1)
(
u
×
v
)
·
u
= (

2
,

5
,
1)
·
(1
,
0
,
2) = 0
,
(
u
×
v
)
·
v
= (

2
,

5
,
1)
·
(

1
,
1
,
3) = 0
.
Theorem 3.5.2 (Properties of Cross Product):
If u,v and w are any vectors
in 3space and k is any scalar, then
(a)
u
×
v
=

(
v
×
u
)
(b)
u
×
(
v
+
w
) = (
u
×
v
) + (
u
×
w
)
(c)
(
u
+
v
)
×
w
= (
u
×
w
) + (
v
×
w
)
(d)
k
(
u
×
v
) = (
ku
)
×
v
=
u
×
(
kv
)
1
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(e)
u
×
0 = 0
×
u
= 0
(f)
u
×
u
= 0
Geometric Interpretation of Cross Product and Determinants:
Theorem 3.5.3 (Area of a Parallelogram)
If u and v are vectors in 3space,
then

u
×
v

is equal to the area of the parallelogram determined by u and v.
Why

u
×
v

=

u

v

sin (
θ
)
?
From Theorem 3.5.1 we have Lagrange’s Identity

u
×
v

2
=

u

2

v

2

(
u
·
v
)
2
and
we have
cos (
θ
) =
u
·
v

u

v

⇒
u
·
v
=

u

v

cos (
θ
)
. Then

u
×
v

2
=

u

2

v

2

(
u
·
v
)
2
=

u

2

v

2
 
u

2

v

2
cos
2
(
θ
)
=

u

2

v

2
(1

cos
2
(
θ
))
=

u

2

v

2
sin
2
(
θ
)
⇒ 
u
×
v

=

u

v

sin (
θ
)
Theorem 3.5.4:
(a) The absolute value of the determinant
det
u
1
u
2
v
1
v
2
is equal to the area of the par
allelogram in 2space determined by the vectors
u
= (
u
1
, u
2
)
and
v
= (
v
1
, v
2
)
.
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 Spring '11
 Zhang
 Equations, Parametric equation

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