
In this section, I will explain what the cross product is, and why the magnitude of the cross product of a and b is a b sin θ, where θ is the angle between a and b.
The cross product of a and b, denoted a×b, is the determinant of the matrix
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i j k a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} ú
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where i, j, and k are the orthogonal unit vectors.
For one thing, this definition implies that i�j=k. (Why? Because if i=[1,0,0] and j=[0,1,0] so i�j is the determinant of
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i j k 1 0 0 0 1 0 ú
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ú= k
That's why.)
So a�b = [a_{2}b_{3}a_{3}b_{2}, a_{3}b_{1}a_{1}b_{3}, a_{1}b_{2}a_{2}b_{1}].
a�b² = (a_{2}b_{3}a_{3}b_{2})² + ( a_{3}b_{1}a_{1}b_{3})² + ( a_{1}b_{2}a_{2}b_{1})²
a�b² = (a_{2}²b_{3}²2a_{2}b_{3}a_{3}b_{2}+a_{3}²b_{2}²) + ( a_{3}²b_{1}²2a_{3}b_{1}a_{1}b_{3}+a_{1}²b_{3}²) + ( a_{1}²b_{2}²2a_{1}b_{2}a_{2}b_{1}+a_{2}²b_{1}²)
a�b² = a_{1}²b_{2}² + a_{1}²b_{3}² + a_{2}²b_{1}² + a_{2}²b_{3}² + a_{3}²b_{1}² + a_{3}²b_{2}²  2a_{1}b_{1} a_{2}b_{2}  2 a_{1}b_{1} a_{3}b_{3}  2 a_{2}b_{2} a_{3}b_{3}
a�b² = a_{1}²b_{1}² + a_{1}²b_{2}² + a_{1}²b_{3}²
+ a_{2}²b_{1}² + a_{2}²b_{2}² + a_{2}²b_{3}²
+ a_{3}²b_{1}² + a_{3}²b_{2}² + a_{3}²b_{3}²
+
a_{1}²b_{1}²
 a_{2}²b_{2}²  a_{3}²b_{3}²  2a_{1}b_{1}a_{2}b_{2}
 2a_{1}b_{1}
a_{3}b_{3}  2a_{2}b_{2}a_{3}b_{3}
a�b² = (a_{1}² + a_{2}² + a_{3}²) (b_{1}² + b_{2}² + b_{3}²)  (a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3})²
a�b² = a² b²  (a·b)²  This is called the Lagrange Identity.
Assuming a, b ≠ 0,
a�b² = a² b²  (a b cos θ)²
a�b² = a² b²(1  cos²θ)
where θ is the angle between a and b. (See "Dot Product" to understand why this is true.)
a�b² = a² b²(sin²θ)
a�b = a b (sin θ)
A geometric interpretation of the magnitude of the cross product is the area of a parallelogram. If vector b represents the base of the parallelogram, and θ is the angle between a and b, then its height is asin θ. Or, if vector a represents the base, then the height is bsin θ. Either way, the parallelogram's area is a b (sin θ).
The area of a triangle formed by two vectors is exactly half the area of the parallelogram formed by those same vectors.
So the area of the triangle is (1/2) a b (sin θ).
Commutative: not quite; interchanging the vectors negates the cross product:
a�b = b�a
(This property is called "anticommutative".)
Associative: no
(i�i)�j = 0�j = 0;
i�(i�j) = i�k = j
Distributive: yes
(a+b)�c = a�c + b�c
a�(b+c) = a�b + a�c
Triple Product
(a�b)·c is a scalar representing the "signed volume" of a parallelepiped (a 6faced polyhedron all of whose faces are parallelograms lying in pairs of parallel planes) whose dimensions and angles are given by the three vectors a, b, and c. If the vectors represent three edges of the solid that meet at a point, and they are named in counterclockwise order from the point of view of the interior of the solid (i j k order, if you will), then the volume will be positive. Otherwise negative.
The triple product is, in its way, commutative. That is, any ordering of a, b, and c give a result with the same absolute value, with its sign determined by the order the edges are named.
Lagrange's formula, sometimes called the "CAB Minus BAC" identity
a�(b�c) = (c�a)b  (b�a)c
Proof:
Let a = [a_{1},a_{2},a_{3}], and let b and c be defined similarly.
b�c = [b_{2}c_{3}b_{3}c_{2}, b_{3}c_{1}b_{1}c_{3}, b_{1}c_{2}b_{2}c_{1}]
a�(b�c) = [a_{2}(b_{1}c_{2}b_{2}c_{1})a_{3}(b_{3}c_{1}b_{1}c_{3}), a_{3}(b_{2}c_{3}b_{3}c_{2})a_{1}(b_{1}c_{2}b_{2}c_{1}), a_{1}(b_{3}c_{1}b_{1}c_{3})a_{2}(b_{2}c_{3}b_{3}c_{2})]
a�(b�c) = [a_{2}b_{1}c_{2}a_{2}b_{2}c_{1}a_{3}b_{3}c_{1}+a_{3}b_{1}c_{3}, a_{3}b_{2}c_{3} a_{3}b_{3}c_{2}a_{1}b_{1}c_{2}+a_{1}b_{2}c_{1}, a_{1}b_{3}c_{1} a_{1}b_{1}c_{3}a_{2}b_{2}c_{3}+a_{2}b_{3}c_{2}](c�a)b = [(a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3})b_{1}, (a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3})b_{2}, (a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3})b_{3}]
(c�a)b = [a_{1}b_{1}c_{1}+a_{2}b_{1}c_{2}+a_{3}b_{1}c_{3}, a_{1}b_{2}c_{1}+a_{2}b_{2}c_{2}+a_{3}b_{2}c_{3}, a_{1}b_{3}c_{1}+a_{2}b_{3}c_{2}+a_{3}b_{3}c_{3}](b�a)c = [(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})c_{1}, (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})c_{2}, (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})c_{3}]
(b�a)c = [a_{1}b_{1}c_{1}+a_{2}b_{2}c_{1}+a_{3}b_{3}c_{1}, a_{1}b_{1}c_{2}+a_{2}b_{2}c_{2}+a_{3}b_{3}c_{2}, a_{1}b_{1}c_{3}+a_{2}b_{2}c_{3}+a_{3}b_{3}c_{3}](c�a)b  (b�a)c = [a_{2}b_{1}c_{2}a_{2}b_{2}c_{1}+a_{3}b_{1}c_{3}a_{3}b_{3}c_{1}, a_{1}b_{2}c_{1}a_{1}b_{1}c_{2}+a_{3}b_{2}c_{3}a_{3}b_{3}c_{2}, a_{1}b_{3}c_{1}a_{1}b_{1}c_{3}+a_{2}b_{3}c_{2}a_{2}b_{2}c_{3}]
From the expansion, we see that a�(b�c) is identical to (c�a)b  (b�a)c
Some advice:
It's hard to remember what dots with what in this formula, so here's a tip. Think about the physical representation of b�c  it's a vector that's normal (that means perpendicular) to the plane defined by vectors b and c. Think of a flagpole sticking straight out of a vast bc plane. When you cross any other vector with that flagpole, the result is perpendicular to the flagpole; in other words the result is in that vast bc plane. So the vector that results from a�(b�c) is in the bc plane, which means it's a linear combination of b and c.
Now look at the formula again. a�(b�c) = (c�a)b  (b�a)c  Do you see how this formula reminds you that a�(b�c) is a linear combination of b and c? Now you need to remember just a few more things, and they should be fairly intuitive:
 The scalar factor for b is the dot product of the other two vectors, and likewise for c.
 This identity has a subtraction in it, not an addition. That should be obvious if you consider what if b=c? Clearly b�c = 0, so a�(b�c) has to be zero, too.
 Use the righthand rule to figure out whether to subtract (c�a)b  (b�a)c or vice versa.
If you follow that advice, you should be able to discern the very similar formula for (a�b)�c  then read on in this page to see if you were right.
a�(b�c) + b�(c�a) + c�(a�b) = 0
Proof:
Add the "CAB Minus BAC" identities:
a�(b�c) = (c�a)b  (b�a)c
b�(c�a) = (a�b)c  (c�b)a
c�(a�b) = (b�c)a  (a�c)b
What about (a�b)�c ?
(a�b)�c = (c�a)b  (b�c)a
Proof:
Let a = [a_{1},a_{2},a_{3}], and let b and c be defined similarly.
a�b = [a_{2}b_{3}a_{3}b_{2}, a_{3}b_{1}a_{1}b_{3}, a_{1}b_{2}a_{2}b_{1}]
(a�b)�c = [(a_{3}b_{1}a_{1}b_{3})c_{3}(a_{1}b_{2}a_{2}b_{1})c_{2}, (a_{1}b_{2}a_{2}b_{1})c_{1}(a_{2}b_{3}a_{3}b_{2})c_{3}, (a_{2}b_{3}a_{3}b_{2})c_{2}(a_{3}b_{1}a_{1}b_{3})c_{1}]
(a�b)�c = [a_{3}b_{1}c_{3}a_{1}b_{3}c_{3}a_{1}b_{2}c_{2}+a_{2}b_{1}c_{2}, a_{1}b_{2}c_{1}a_{2}b_{1}c_{1}a_{2}b_{3}c_{3}+a_{3}b_{2}c_{3}, a_{2}b_{3}c_{2}a_{3}b_{2}c_{2}a_{3}b_{1}c_{1}+a_{1}b_{3}c_{1}]This is equal to (c�a)b  (b�c)a, which you can verify using the same method as in the Lagrange formula, above.
Some advice:
In both of these identities, you have (c�a)b as a positive number, with the other vector subtracted from this one. Look at them again:
a�(b�c) = (c�a)b  (b�a)c  because the resulting vector is in the bc plane
(a�b)�c = (c�a)b  (b�c)a  because the resulting vector is in the ab planeCall me a cab!
(OK, you're a cab!)
The other Jacobi Identity
(a�b)�c + (b�c)�a + (c�a)�b = 0
Proof:
Add the "CAB Minus BCA" identities:
(a�b)�c = (c�a)b  (b�c)a
(b�c)�a = (a�b)c  (c�a)b
(c�a)�b = (b�c)a  (a�b)c
Cross Product, from the Fact Index
http://steiner.math.nthu.edu.tw/disk3/cal03/html/vector/vector.html
Triple Product  a·(b�c) is a scalar representing the "signed volume" of a parallelepiped
Triangle Area Using Vectors, part 1
Triangle Area using Vectors, part 2
Triangle Area using Determinant
Geometry and Trigonometry, and in particular, the Points and Lines section.
The webmaster and author of this Math Help site is Graeme McRae.