Section 2.3. The Cross Product
Recall that a vector can be uniquely determined by its length and direction.
De…nition. The cross product of two vectors ~u and ~v; denote by ~u £ ~v;
is a vector with length
j~u £ ~vj = j~uj j~v j sin µ
(µ is the angle between ~u and ~v; 0 · µ · ¼)
and the direction that is perpendicular to both ~u and ~v following the righthand rule: Suppose that ~u and ~v lie on the screen as you see. Then ~u £ ~v is
perpendicular to the screen and is pointing towards you.
W=U xV
V
U
The cross product is also referred to as Outer Product or Vector Product.
Example 3.1. (a) ~i £ ~j = ~k; (b) ~j £ ~k = ~i; (c) ~k £ ~i = ~j:
1
In physics, the measurement for turning e¤ect, torque, is de…ned as the
~ and force vector F~ :
cross product of length vector R
¯
¯ ¯ ¯ ³¯ ¯
´
³
´
¯~ ~¯ ¯~¯ ¯~¯
¯R £ F ¯ = ¯R¯ ¯F ¯ sin µ = (length of the wrench)£ perpendicular component of F~
T= R x F
F
|F| sin θ
F
When using a wrench to tighten a screw, the
strength is measured by Torque
T=| R x F |
θ
R
Example 3.2. A bolt is tightened by applying 40 Newton force to a
0:025m wrench at 75degree angle. Find the magnitude of torque.
Solution: Let F~ be the force, ~r be length vector. Then
¯ ¯
¯~¯
¯F ¯ = 40; j~rj = 0:025; µ = 750 :
¯
¯ ¯ ¯
¯~
¯ ¯ ¯
¯F £ ~r¯ = ¯F~ ¯ j~rj sin µ = 40 £ 0:025 £ sin 75o = 9:66 (J = N m)
Geometrically, let ~u and ~v be two generating edges of a parallelogram as
shown in the following …gure.
2
V
V
h = |V|sinθ
θ
θ
U
U
The area of a parallelogram is
Area = length of the base £ height
= j~uj h = j~uj j~v j sin µ
= j~u £ ~vj :
The area of a triangle is
Area =
area of parallelogram
j~u £ ~vj
=
:
2
2
Property of Cross product:
(1) ~u £ ~v ? ~u; ~v
This is by de…nition.
(2) ~u £ ~v = ~0 i¤ ~u == ~v (parallel)
This is due to the fact that ~u == ~v i¤ sin µ = 0:
(3) ~u £ ~v = ¡~v £ ~u
This is because the right-hand rule.
3
V
W= U x V
V
U
VxU
U
(4) (¸~u) £ ~v = ¸(~u £ ~v) = ~u £ (¸~v)
(5) ~u £ (~v + w
~ ) = ~u £ ~v+ ~u £ w
~
Example 3.3. Let ~u =< 1; 3; 4 >; ~v =< 2; 7; ¡5 > : Find ~u £ ~v:
Solution: We shall use the facts that (From Example 3.1)
~i £ ~j = ~k; ~j £ ~k = ~i; ~k £ ~i = ~j;
and above properties.
³
´ ³
´
~
~
~
~
~
~
~u £ ~v = i + 3j + 4k £ 2i + 7j ¡ 5k
³
´
= ~i £ 2~i + 7~j ¡ 5~k
³
´
+ 3~j £ 2~i + 7~j ¡ 5~k
³
´
~
~
~
~
+ 4k £ 2i + 7j ¡ 5k
³ ´
³ ´
³ ´
= ~i £ 2~i + ~i £ 7~j ¡ ~i £ 5~k
³ ´
³ ´
³ ´
+ 3~j £ 2~i + 3~j £ 7~j ¡ 3~j £ 5~k
³ ´
³ ´
³ ´
~
~
~
~
~
+ 4k £ 2i + 4k £ 7j ¡ 4k £ 5~k
= 7~k + 5~j
¡ 6~k ¡ 15~i
+ 8~j ¡ 28~i
= ¡43~i + 13~j + ~k:
4
In general, let ~u =< x1 ; y1 ; z1 >; ~v =< x2 ; y2 ; z2 > :Then
³
´ ³
´
~u £ ~v = x1~i + y1~j + z1~k £ x2~i + y2~j + z2~k
= (y1 z2 ¡ y2 z1 )~i + (z1 x2 ¡ z2 x1 ) ~j + (x1 y2 ¡ x2 y1 ) ~k:
Cross Product in Component Form: We (some of us) may already
know that 3 £ 3 determinants can be computed through the expansion:
¯
¯
¯ a b c ¯
¯
¯
¯
¯
¯
¯
¯
¯
¯ y1 z1 ¯
¯ x1 z1 ¯
¯ x1 y1 ¯
¯ x1 y1 z 1 ¯ = a ¯
¯
¯
¯
¯
¯
¯
¯
¯ y2 z2 ¯ ¡ b ¯ x2 z2 ¯ + c ¯ x2 y2 ¯ :
¯ x2 y2 z 2 ¯
For instance,
¯
¯ 1 2 3
¯
¯ 1 3 4
¯
¯ 2 7 ¡5
¯
¯
¯
¯
¯
¯= 1¢¯ 3 4
¯
¯ 7 ¡5
¯
¯
¯
¯
¯
¯ ¡2¯ 1 4
¯
¯ 2 ¡5
¯
¯
¯
¯
¯+3¯ 1 3
¯
¯ 2 7
¯
¯
¯
¯
= (¡15 ¡ 28) ¡ 2 (¡5 ¡ 8) + 3 (7 ¡ 6) = ¡14:
We …nd the cross product formula may be memorized by using the determinant expansion formula:
¯
¯
¯ ~i ~j ~k ¯ ¯
¯
¯
¯
¯
¯
¯
¯ ¯ y1 z 1 ¯
¯ x1 z1 ¯
¯ x1 y1 ¯
¯~i ¡ ¯
¯~ ¯
¯~
~u £ ~v = ¯¯ x1 y1 z1 ¯¯ = ¯¯
¯
¯ x2 z2 ¯ j + ¯ x2 y2 ¯ k:
y
z
2
2
¯ x2 y2 z2 ¯
Example 3.4. Find ~u £ ~v if (a) ~u =< 1; 3; 4 >; ~v =< 2; 7; ¡5 >; (b)
~u =< ¡1; 2; 1 >; ~v =< 1; ¡2; 0 > :
Solution: (a) Note ~u and ~v are the same as in Example 3.3.
¯
¯ ¯
¯
¯ ~i ~j ~k ¯ ¯ ~i ~j ~k ¯
¯
¯ ¯
¯
~u £ ~v = ¯¯ x1 y1 z1 ¯¯ = ¯¯ 1 3 4 ¯¯
¯ x2 y2 z2 ¯ ¯ 2 7 ¡5 ¯
¯
¯
¯
¯
¯
¯
¯ 3 4 ¯
¯ 1 4 ¯
¯ 1 3 ¯
¯
¯
¯
¯
¯
¯ ~k
~
~
=¯
i¡¯
j+¯
7 ¡5 ¯
2 ¡5 ¯
2 7 ¯
= (¡15 ¡ 28)~i ¡ (¡5 ¡ 8) ~j + (7 ¡ 6) ~k
= ¡43~i + 13~j + ~k:
5
(b)
¯
¯
¯
~u £ ~v = ¯¯
¯
¯
¯
= ¯¯
¯
~i
~j ~k ¯
¯
¡1 2 1 ¯¯
1 ¡2 0 ¯
¯
¯
2 1 ¯¯~ ¯¯ ¡1 1
i¡¯
¡2 0 ¯
1 0
= 2~i + ~j:
¯
¯
¯
¯
¯ ~j + ¯ ¡1 2
¯
¯ 1 ¡2
¯
¯
¯ ~k
¯
Example 3.5. Consider the triangle with vertices P (1; 4; 6) ; Q (¡2; 5; ¡1) ;
R (1; ¡1; 1) :
Q
V
P
R
U
(a) Find a vector that is perpendicular to this triangle.
(b) Find the area of the triangle.
Solution: (a) Let
¡! ¡! ¡!
~u = P R = OR ¡ OP = h1; ¡1; 1i ¡ h1; 4; 6i = h0; ¡5; ¡5i
¡! ¡! ¡!
~v = P Q = OQ ¡ OP = h¡2; 5; ¡1i ¡ h1; 4; 6i = h¡3; 1; ¡7i
Then the vector ~u £ ~v is a vector perpendicular to both ~u and ~v , and consequently it is perpendicular to the triangle. Now,
¯
¯
¯ i
¯ ¯
¯
¯
¯
¯
¯
j
k
¯
¯ ¯ ¡5 ¡5 ¯
¯ 0 ¡5 ¯
¯ 0 ¡5 ¯
¯
¯
¯
¯
¯
¯
¯
¯~
~
~
~u £ ~v = ¯ 0 ¡5 ¡5 ¯ = ¯
¯ i ¡ ¯ ¡3 ¡7 ¯ j + ¯ ¡3 1 ¯ k
1
¡7
¯ ¡3 1 ¡7 ¯
³
´
= 40~i + 15~j ¡ 15~k = 5 8~i + 3~j ¡ 3~k :
6
Ans: h8; 3; ¡3i is perpendicular to the triangle.
(b)
´¯
1
1 ¯¯ ³
¯
Area = j~u £ ~vj = ¯5 8~i + 3~j ¡ 3~k ¯
2¯
2
´¯
5 ¯³ ~
¯
= ¯ 8i + 3~j ¡ 3~k ¯
2
p
5p
5 82
=
64 + 9 + 9 =
= 22: 638:
2
2
Triple products. There are at least two ways to de…ne product for 3
vectors ~u; ~v and w:
~
(i) (scalar) triple product: (~u £ ~v) ¢ w
~ is a scalar.
Ux V
W
h
θ
V
U
Consider the parallelepiped formed by three vectors ~u; ~v and w:The
~
bottom face is a parallelogram by ~u and ~v; whose area is j~u £ ~vj : The cross
product (~u £ ~v ) is perpendicular to the bottom face, and
j(~u £ ~v ) ¢ wj
~ = j~u £ ~vj (jw
~ cos µj) = j~u £ ~v j h = volume of parallelepiped
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If ~u =< x1 ; y1 ; z1 >; ~v =< x2 ; y2 ; z2 >; w
~ =< x3 ; y3 ; z3 >; then
¯
¯
¯ ~i ~j ~k ¯
¯
¯
(~u £ ~v ) ¢ w
~ = ¯¯ x1 y1 z1 ¯¯ ¢ w
~
¯ x2 y2 z2 ¯
¯
¯
¯
¯
¯ ¶
µ¯
¯ y1 z1 ¯
¯ x1 z1 ¯
¯ x1 y1 ¯
¯~i ¡ ¯
¯~ ¯
¯~
= ¯¯
¯ x2 z2 ¯ j + ¯ x2 y2 ¯ k ¢ < x3 ; y3 ; z3 >
y2 z2 ¯
¯
¯
¯
¯
¯
¯
¯ y1 z 1 ¯
¯ x1 z1 ¯
¯ x1 y1 ¯
¯x ¡ ¯
¯y + ¯
¯z
= ¯¯
y2 z2 ¯ 3 ¯ x2 z2 ¯ 3 ¯ x2 y2 ¯ 3
¯
¯
¯
¯ ¯
¯
¯ x3 y3 z3 ¯
¯ x1 y1 z1 ¯ ¯ x1 y1 z1 ¯
¯
¯
¯
¯ ¯
¯
= ¯¯ x1 y1 z1 ¯¯ = ¡ ¯¯ x3 y3 z3 ¯¯ = ¯¯ x2 y2 z2 ¯¯ :
¯ x2 y2 z2 ¯
¯ x2 y2 z2 ¯ ¯ x3 y3 z3 ¯
Example 3.6. Show that the following three vectors
~u =< 1; 4; ¡7 >; ~v =< 2; ¡1; 4 >; w
~ =< 0; ¡9; 18 >
are on the same plane.
Solution:
¯
¯
¯ 1 4 ¡7 ¯
¯
¯
(~u £ ~v) ¢ w
~ = ¯¯ 2 ¡1 4 ¯¯
¯ 0 ¡9 18 ¯
= 1 (¡1) 18 + 4 ¢ 4 ¢ 0 + (¡7) 2 (¡9) ¡ (¡7) (¡1) 0 ¡ 4 ¢ 2 ¢ 18 ¡ 1 ¢ 4 ¢ (¡9)
= ¡18 + 126 ¡ 144 + 36 = 36 = 0:
Since the volume of the parallelepiped is zero, they are on one plane.
(ii) Another way to de…ne triple product is called vector triple product:
(~u £ ~v) £ w
~ is a vector. One can show the following formula
³
´
³
´
~
~
~
~a £ b £ ~c = (~a ¢ ~c) b ¡ ~a ¢ b ~c
Homework:
1. Which of the following expressions are meaningful?
³
´
(a) ~a £ ~b £ ~c
8
³
´
(b) ~a £ ~b ¢ ~c
³
´
(c) ~a £ ~b + ~c
³
´
(d) ~a ¢ ~b ~c
³
´ ³
´
(e) ~a ¢ ~b £ ~c ¢ d~
³
´ ³
´
~
~
(f) ~a £ b ¢ ~c £ d
¯ ¯
¯ ¯
2. Vector ~a and ~b are shown below. Given j~aj = 3; ¯~b¯ = 2:
¯
¯
¯
¯
(a) Find ¯~a £ ~b¯
(b) ~a £~b = hx; y; zi has three components x; y; z: For each component,
determine whether it is positive, or zero, or negative.
Z
r
b
Y
r
a
X
³
´
(c) Find ~a £ ~b ¢ ~k; where ~k = h0; 0; 1i :
3. Find a unit vector that is orthogonal to the plane passing through three
points P; Q; and R: Find also the triangle whose vertices are P; Q; and
R:
(a) P (0; ¡1; 0) ; Q (3; 1; ¡2) ; R (4; 3; 1)
9
(b) P (2; ¡1; 1) ; Q (1; 2; ¡3) ; R (1; ¡3; 2)
4. Determine whether four points P; Q; R; S lie on a plane. If not, …nd
the volume of the parallelepiped with edges P Q; P R; P S:
(a) P (0; ¡1; 0) ; Q (3; 1; ¡2) ; R (4; 3; 1) ; S (2; 2; ¡1)
(b) P (1; 1; 1) ; Q (2; 6; ¡1) ; R (4; 0; 1) ; S (6; 10; ¡3)
5. Find the magnitude of the torque about P if a 36 lb force is applied as
shown.
4 ft
P
4 ft
30 o
36 lb
6. Determine whether the following formulas hold:
³
´ ³
´
³
´
~
~
~
(a) ~a ¡ b £ ~a + b = 2 ~a £ b
³
´ ³
´
³
´
(b) ~a ¡ ~b £ ~a ¡ ~b = ¡2 ~a £ ~b
³
´ ³
´
(c) ~a ¡ ~b ¢ ~a + ~b = 0
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