Instructor: Longfei Li
Math 243 Lecture Notes
12.4 The Cross Product
Another way to multiply vectors: the cross product
Definition: The cross product of a =< a1 , a2 , a3 > and b =< b1 , b2 , b3 > is the vector
a × b =< a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 >
Remark: The cross product is defined only for 3D vectors.
Remark: The cross product a × b is a vector; the dot product a · b is a scalar (real number); the
scalar multiplication cv is a vector.
We can write the definition of cross product in terms of determinant.
a × b =< a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 >
a2 a3 a1 a3 a1 a2 k
=
i−
j+
b2 b3 b1 b3 b1 b2 i
j k = a1 a2 a3 b1 b2 b3 The determinant form is easier to remember.
Example: Evaluate a × b
i
a × b = 1
2
for a =< 1, 0, 1 >, b =< 2, 3, 1 >
j k
0 1 =< 0(1) − 3(1), 2(1) − 1(1), 1(3) − 2(0) >=< −3, 1, 3 >
3 1
Calculate a × a = 0 for any a
i
j
k
a × a = a1 a2 a3 =< a2 a3 − a2 a3 , a1 a3 − a1 a3 , a1 a2 − a1 a2 >= 0
a1 a2 a3 Theorem: The vector a × b is orthogonal to both a and b
Proof:
a · (a × b) =< a1 , a2 , a3 > · < a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 >
= a1 a2 b3 − a1 a3 b2 + a2 a3 b1 − a2 a1 b3 + a3 a1 b2 − a3 a2 b1
=0
So a × b is orthogonal to a. Similarly, a × b is orthogonal to b.
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Direction of a × b: determined by the Right-hand Rule
Example:
i j k
i × j = 1 0 0 =< 0, 0, 1 >= k
0 1 0 i j k
j × i = 0 1 0 =< 0, 0, −1 >= −k
1 0 0 Similarly,
j × k = i, k × j = −i
k × i = j, i × k = −j
In general, a × b 6= b × a not commutative!
Theorem: If θ is the angle between a and b (so 0 ≤ θ ≤ π), then
|a × b| = |a||b| sin θ
Proof:
|a × b|2 = (a2 b3 − a3 b2 )2 + (a3 b1 − a1 b3 )2 + (a1 b2 − a2 b1 )2
= a22 b23 − 2a2 b3 a3 b2 + a23 b22 + a23 b21 − 2a3 b1 a1 b3 + a21 b23 + a21 b22 − 2a1 b2 a2 b1 + a22 b21
= (a21 + a22 + a23 )(b21 + b22 + b23 ) − (a1 b1 + a2 b2 + a3 b3 )2
= |a|2 |b|2 − (a · b)2
= |a|2 |b|2 − |a|2 |b|2 cos2 θ
= |a|2 |b|2 sin2 θ
Since sin θ ≥ 0 for 0 ≤ θ ≤ π, we have
⇒ |a × b| = |a||b| sin θ
Corollary: Two nonzero vectors a and b are parallel if and only if
a×b=0
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The geometric understanding: |a × b| is equal to the area of the parallelogram determined by a and b
Properties:
1. a × b = -a × b
2. (ca) × b = c(a × b) = a × (cb)
3. a × (b + c) = a × b + a × c
4. (a + b) × c = a × c + b × c
5. a · (b × c) = (a × b) · c
6. a × (b × c) = (a · c)b − (a · b)c
The properties are proved using the definition.
a · (b × c) = a1 (b2 c3 − b3 c2 ) + a2 (b3 c1 − b1 c3 ) + a3 (b1 c2 − b2 c1 )
= a1 b2 c3 − a1 b3 c2 + a2 b3 c1 − a2 b1 c3 + a3 b1 c2 − a3 b2 c1
= (a2 b3 − a3 b2 )c1 + (a3 b1 − a1 b3 )c2 + (a1 b2 − a2 b1 )c3
= (a × b) · c
Prove the rest as exercises.
Triple Product:
a1 a2 a3 a · (b × c) = b1 b2 b3 c1 c2 c3 3
Geometric Interpretation: |a · (b × c)| equals the volume of the parallelepiped determined by the
vectors a, b and c
Proof: V = Ah = |b × c||a| cos θ = |a · (b × c)|.
Coplanar: If a · (b × c) = 0, then they lie in the same plane; i.e. they are coplanar.
Example: Show a =< 1, 4, −7 >, b =< 2, −1, 4 > and c =< 0, −9, 18 > are coplanar.
Solution:
1 4 −7
a · (b × c) = 2 −1 4 0 −9 18 −1 4 2 4 2 −1
= 1
− 4
− 7
−9 18
0 18
0 −9
= 1(18) − 4(36) − 7(−18) = 0
So a, b and c are coplanar.
Application of the Cross Product:
τ =r×F
here τ is the torque, r is the position vector and F is the force.
Example: Tighten a bolt with a wrench
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