Download 1.1.3. Vector product (a.k.a cross product) Given two vectors A and B

1.1.3. Vector product (a.k.a cross product)
Given two vectors A and B. We define the vector product of A and B to be
a vector C:
C=A×B
where
1. The length of C is the area of the parallelogram spanned by A and B:
|C| = |A||B|| sin(A, B)|;
2. The direction of C is perpendicular to the plane formed by A and B. And
the three vectors A, B, and C follow the right-hand rule.
The right-hand rule is: when the four fingers of the right hand turn from A to
B, the thumb points to the direction of C.
(Figure 1.1.3.1. Definition of cross product.)
Right hand rule
C=AxB
B
Area
A
Figure 1.1.3.1. Definition of cross product.
Properties:
A × B = −B × A;
A × (B + C) = A × B + A × C;
AkB is the same as A × B = 0
where the symbol k means parallel.
Example 1.1.3a: Recall the three vectors i1 , i2 , i3 . They follow the right hand
rule. By definition we can find that
i1 × i1 = 0,
i2 × i2 = 0,
i3 × i3 = 0;
i1 × i2 = i3 ,
i 2 × i3 = i1 ,
i 3 × i1 = i2 .
and
With Example 1.1.3a and the distributive property, we can find a formula for
the product. Let
A = a1 i1 + a2 i2 + a3 i3 ,
B = b 1 i1 + b 2 i2 + b 3 i3 .
Then (please do it on your own. you can do it.)
i1
A × B = a1
b1
i2
a2
b2
i3 a3 .
b3 Example 1.1.3b. An electric charge e moving with velocity v in a magnetic
field H experiences a force:
e
F = (v × H)
c
where c is the speed of light.
Example 1.1.3c. The moment M of a force F about a point O is
M = r × F.
(Figure 1.1.3.2. Moment of a force.)
M
r
O
F
Figure 1.1.3.2. The moment of a force.
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1.1.4. Product of three vectors.
Given three vectors A, B, and C. We list three products with formula
(A × B) × C = B(A · C) − A(B · C);
A × (B × C) = B(A · C) − C(A · B);
a1
(A × B) · C = b1
c1
a2 a3 b2 b3 c2
c3 where the entries are the coordinates of the three vectors. The first two products
are called vector triple products, the third is called scalar triple product. The proof
for the formulas for the vector triple products are complicated. But the proof for
the formula for the scalar triple product is straightforward. The reader should be
able to do it alone.
To remember the formulas for the two vector triple products, there is a quick way.
You see that the final product of the first vector triple product will be perpendicular
to A× B, so it will lie in the plane spanned by A and B. It is perpendicular to C, so
there will be no component in the C direction. So the first vector triple product is
a linear combination of A and B, not C. The coefficients are the inner products of
the remaining two vectors, with a minus sign for the second term; while the middle
vector B is the first term.
Recall that the magnitude (length) of A × B is the area of the parallelogram
spanned by A and B, and the inner product with C is this magnitude times |C| cos φ,
which is exactly the height of the parallelepiped with a “slanted height” |C| and a
bottom parallelogram spanned by A and B. Thus the magnitude of the scalar triple
product is the volume of the parallelepiped formed by the three vectors. See Figure
1.1.3.3.
(Figure 1.1.3.3. Volume of the parallelepiped formed by three vectors.)
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AxB
height =
C
projection
B
Area
of C.
A
Figure 1.1.3.3. The volume of the parallelepiped is the magnitude of (AxB) \centerdot C.
One can form other triple products, but they all can be reduced quickly to one
of the three mentioned here. One may notice that the second vector triple product
can be reduced to the first vector product easily. So essentially there is only one
vector triple product and one scalar triple product.
1.2. Variable vectors.
1.2.1. Vector functions of a scalar argument.
(−∞ < t < ∞).
Example 1.2.1a. A(t) = (cos t, sin t, 0),
The graph (the collection of all the tips of the vector A(t)) is a circle.
Example 1.2.1b. A(t) = (cos t, sin t, t),
(−∞ < t < ∞).
The graph is a helix.
Example 1.2.1c. A(t) = (3t, 2t, −t) = (3, 2, −1)t. It represents a straight line.
The general formula for a straight line is
A(t) = tα + β
where α and β are numerical vectors independent of t.
1.2.2. The derivatives of a vector function.
Let
A(t) = (A1 (t), A2 (t), A3 (t)) = A1 (t)i1 + A2 (t)i2 + A3 (t)i3 .
Then
dA(t)
dA2 (t)
dA3 (t)
dA1 (t) dA2 (t) dA3 (t)
dA1 (t)
=(
,
,
)=
i1 +
i2 +
i3 .
dt
dt
dt
dt
dt
dt
dt
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Formal definition:
A(t + ∆t) − A(t)
.
∆t→0
∆t
A0 (t) = lim
(Figure 1.2.1. Derivative: pay attention to the direction of the difference quotient.)
A (t)
A (t+ dt)
Figure 1.2.1. Derivative: tangent direction.
Example 1.2.2a. Let r(t) be the position vector of a moving particle. Then
dr(t)
= v(t)
dt
is the velocity;
d2 r(t)
dv(t)
=
= a(t)
2
dt
dt
is the acceleration.
Example 1.2.2b. Consider r(t) = (cos t, sin t). Then r0 (t) = (− sin t, cos t).
Note that the derivative is tangent to the graph of r(t). Consider R(t) = 2(cos t, sin t).
Then R0 (t) = 2(− sin t, cos t). Notice that the magnitude of the derivative is twice
as large. See Figure 1.2.2.
(Figure 1.2.2. Derivative: direction and magnitude.)
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R’
r’
R
r
1
2
Figure 1.2.2. Derivative: direction and magnitudes.
Simple rules:
d
dt (A ± B)
d
dt (cA)
d
dt (A · B)
d
dt (A × B)
=
=
=
=
d
d
dt A ± dt B,
dc
d
dt A + c dt A,
dA
dB
dt · B + A · dt .
dA
dB
dt × B + A × dt .
(1)
1.2.3. The integral of a vector function.
The integral of a vector function is defined also component-wise.
Let
A(t) = (A1 (t), A2 (t), A3 (t)) = A1 (t)i1 + A2 (t)i2 + A3 (t)i3 .
Then
Z
B(t) =
Z
A(t) dt = (
Z
A1 (t) dt,
Z
A2 (t) dt,
A3 (t) dt).
Combining Lectures 1 and 2, we have covered this week the following sections
of our text book (Vector and Tensor Analysis with Applications, by Borisenko etc.)
1.2.3, 1.4, 1.5, and 1.7.
End of Lecture 2.
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