Unformatted text preview: 1 4.1 Areas, Volumes, and Cross Products Chapter 4. Determinants
4.1 Areas, Volumes, and Cross Products Note. Area of a Parallelogram.
Consider the parallelogram determined by two vectors a and b: Figure 4.1, Page 239.
Its area is
A = Area = (base) × (height) = a b sin θ
1 − cos2 θ. =ab
Squaring both sides:
A2 = a 2 b 2(1 − cos2 θ) = a 2 b 2 −a 2 b 2 cos2 θ 2 4.1 Areas, Volumes, and Cross Products a = 2 b 2 − (a · b)2. Converting to components a = [a1, a2] and b = [b1, b2] gives
A2 = (a1b2 − a2b1)2
or A = a1b2 − a2b1. Deﬁnition. For a 2 × 2 matrix A = of A as a1 a2
b1 b2 det(A) = a1b2 − a2b1 = , deﬁne the determinant a1 a2
b1 b2 . Example. Page 249 number 26. Deﬁnition. For two vectors b = [b1, b2, b3] and c = [c1, c2, c3] deﬁne the
cross product of b and c as
b×c= b2 b3
c2 c3 ˆ−
i b1 b3
c1 c3 ˆ+
j b1 b2
c1 c2 ˆ
k. 4.1 Areas, Volumes, and Cross Products 3 Note. We can take dot products and ﬁnd that b × c is perpendicular to
both b and c. Note. If b, c ∈ R3 are not parallel, then there are two directions perpendicular to both of these vectors. We can determine the direction of b × c
by using a “right hand rule.” If you curl the ﬁngers of your right hand
from vector b to vector c, then your thumb will point in the direction of
b × c: Figure 4.3, Page 242. Example. Page 248 number 16. 4 4.1 Areas, Volumes, and Cross Products a1 a2 a3 b b b deﬁne the determiDeﬁnition. For a 3 × 3 matrix A = 1 2 3 c1 c2 c3
nant as
a1 a2 a3
det(A) = b1 b2 b3 = a1
c1 c2 c3 b2 b3
c2 c3 − a2 b1 b3
c1 c3 + a3 b1 b2
c1 c2 . Note. We can now see that cross products can be computed using determinants:
ˆˆk
ijˆ
b × c = b1 b2 b3 .
c1 c2 c3 5 4.1 Areas, Volumes, and Cross Products Theorem. The area of the parallelogram determined by b and c is b×c .
Proof. We know from the ﬁrst note of this section that the area squared
is A2 = c b − (c · b)2. In terms of components we have A2 = (c2 + c2 + c2)(b2 + b2 + b2) − (c1b1 + c2b2 + c3b3)2.
1
2
3
1
2
3
Multiplying out and regrouping we ﬁnd that
A2 = b2 b3
c2 c3 2 + b1 b3
c1 c3 2 + b1 b2
c1 c2 Taking square roots we see that the claim is veriﬁed. 2 .
QED Theorem. The volume of a box determined by vectors a, b, c ∈ R3 is
V = a1(b2c3 − b3c2) − a2(b1c3 − b3c1) + a3(b1c2 − b2c1 ) = a · b × c.
Proof. Consider the box determined by a, b, c ∈ R3: Figure 4.5, Page 244. 6 4.1 Areas, Volumes, and Cross Products The volume of the box is the height times the area of the base. The area
of the base is b × c by the previous theorem. Now the height is
h = a  cos θ = b×c a  cos θ b×c = ( b × c ) · a 
b×c . (Notice that if b × c is in the opposite direction as given in the illustration
above, then θ would be greater than π/2 and cos θ would be negative.
Therefore the absolute value is necessary.) Therefore
V = (Area of base)(height) = b × c (c × c) · a
b×c = ( b × c ) · a .
QED Note. The volume of a box determined by a, b, c ∈ R3 can be computed
in a similar manner to cross products:
a1 a2 a3
V =  det(A) = b1 b2 b3
c1 c2 c3 Example. Page 249 number 37. . 7 4.1 Areas, Volumes, and Cross Products Theorem 4.1. Properties of Cross Product.
Let a, b, c ∈ R3.
(1) Anticommutivity: b × c = −c × b
(2) Nonassociativity of ×: a × (b × c) = (a × b) × c (That is, equality
does not in general hold.)
(3) Distributive Properties: a × (b + c) = (a × b) + (a × c)
(a + b) × c = (a × c) + (b × c)
(4) Perpendicular Property: b · (b × c) = (b × c) · c = 0
(5) Area Property: b × c = Area of the parallelogram determined by b
and c
(6) Volume Property: a · (b × c) = (a × b) · c = ±Volume of the box
determined by a, b, and c.
(7) a × (b × c) = (a · c)b − (a · b)c
Proof of (1). We have
b×c = b2 b3
c2 c3 ˆ−
i b1 b3
c1 c3 ˆ
j+ b1 b2
c1 c2 ˆ
k ˆ
i
j
= (b2c3 − b3c2 )ˆ − (b1c3 − b3c1 )ˆ + (b1c2 − b2c1)k
ˆ
i
j
= − (b3c2 − b2c3 )ˆ − (b3c1 − b1c3)ˆ + (b2c1 − b1c2)k 8 4.1 Areas, Volumes, and Cross Products = − c2 c3
b2 b3 ˆ−
i c1 c3
b1 b3 ˆ
j+ c1 c2
b1 b2 ˆ
k = −c × b
QED Example. Page 249 number 56. ...
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 Fall '04
 IgorDolgachev
 Determinant, Vectors, Dot Product, Bivector, Righthand rule

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