Calculus 2nd Edition Rogawski

SECTION 13.4 The Cross Product 697
z
EXAMPLE 5 Velocity in a Magnetic Field The force F on a proton moving at velocity
v m/s in a uniform magnetic field B (in teslas) is F = q(v × B) in newtons, where q =
1.6 × 10 −19 coulombs (Figure 8). Calculate F if B = 0.0004k T and v has magnitude
10 6 m/s in the direction −j + k.
v
B
F
Solution The vector −j + k has length √ 2, and since v has magnitude 10 6 ,
( ) −j + k
v = 10 6 √
2
x
FIGURE 8 A proton in a uniform magnetic
field travels in a helical path.
y
Therefore, the force (in newtons) is
( ) −j + k
F = q(v × B) = 10 6 q √ × (0.0004k) = 400q √ ((−j + k) × k)
2 2
= − 400q √
2
i = −400(1.6 × 10−19 )
√
2
i ≈−(4.5 × 10 −17 )i
Cross Products, Area, and Volume
Cross products and determinants are closely related to area and volume. Consider the
parallelogram P spanned by nonzero vectors v and w with a common basepoint. In
Figure 9(A), we see that P has base b = ∥v∥ and height h = ∥w∥ sin θ, where θ is the
angle between v and w. Therefore, P has area A = bh = ∥v∥∥w∥ sin θ = ∥v × w∥.
z
v × w
w
h
v × w
θ
v
u
h
θ
w
x
(A) The area of the parallelogram P is
|| v × w|| = || v|| || w|| sin θ.
y
(B) The volume of the parallelpiped P is
| u . (v × w)| .
v
FIGURE 9
A “parallelepiped” is the solid spanned by
three vectors. Each face is a parallelogram.
Next, consider the parallelepiped P spanned by three nonzero vectors u, v, w in R 3
[the three-dimensional prism in Figure 9(B)]. The base of P is the parallelogram spanned
by v and w, so the area of the base is ∥v × w∥. The height of P is h = ∥u∥ ·|cos θ|, where
θ is the angle between u and v × w. Therefore,
Volume of P = (area of base)(height) = ∥v × w∥ · ∥u∥ ·|cos θ|
But ∥v × w∥∥u∥ cos θ is equal to the dot product of v × w and u. This proves the formula
Volume of P =|u · (v × w)|
The quantity u · (v × w), called the vector triple product, can be expressed as a
determinant. Let
u = ⟨a 1 ,b 1 ,c 1 ⟩ , v = ⟨a 2 ,b 2 ,c 2 ⟩ , w = ⟨a 3 ,b 3 ,c 3 ⟩