Introduction to Vector Product Identities

Let a be a vector whose components are called [a₁, a₂, a₃], and let b and c be vectors as well.

The cross product a�b = [a₂b₃-a₃b₂, a₃b₁-a₁b₃, a₁b₂-a₂b₁].  The cross product is perpendicular to both a and b, and has a magnitude equal to |a| |b| sin θ, which is the area of a parallelogram, two sides of which are formed by vectors a and b.

The dot product a·b = |a| |b| cos θ, and in three dimensions, a·b = a₁b₁ + a₂b₂ + a₃b₃

The triple product (a�b)·c is a scalar representing the "signed volume" of a parallelepiped.

Lagrange's formula

a�(b�c) = (c�a)b - (b�a)c is Lagrange's formula, sometimes called the "CAB Minus BAC" identity

Proof:

Let a = [a₁,a₂,a₃], and let b and c be defined similarly.

b�c = [b₂c₃-b₃c₂, b₃c₁-b₁c₃, b₁c₂-b₂c₁]

a�(b�c) = [a₂(b₁c₂-b₂c₁)-a₃(b₃c₁-b₁c₃), a₃(b₂c₃-b₃c₂)-a₁(b₁c₂-b₂c₁), a₁(b₃c₁-b₁c₃)-a₂(b₂c₃-b₃c₂)]
a�(b�c) = [a₂b₁c₂-a₂b₂c₁-a₃b₃c₁+a₃b₁c₃, a₃b₂c₃- a₃b₃c₂-a₁b₁c₂+a₁b₂c₁, a₁b₃c₁- a₁b₁c₃-a₂b₂c₃+a₂b₃c₂]

(c�a)b = [(a₁c₁+a₂c₂+a₃c₃)b₁, (a₁c₁+a₂c₂+a₃c₃)b₂, (a₁c₁+a₂c₂+a₃c₃)b₃]
(c�a)b = [a₁b₁c₁+a₂b₁c₂+a₃b₁c₃, a₁b₂c₁+a₂b₂c₂+a₃b₂c₃, a₁b₃c₁+a₂b₃c₂+a₃b₃c₃]

(b�a)c = [(a₁b₁+a₂b₂+a₃b₃)c₁, (a₁b₁+a₂b₂+a₃b₃)c₂, (a₁b₁+a₂b₂+a₃b₃)c₃]
(b�a)c = [a₁b₁c₁+a₂b₂c₁+a₃b₃c₁, a₁b₁c₂+a₂b₂c₂+a₃b₃c₂, a₁b₁c₃+a₂b₂c₃+a₃b₃c₃]

(c�a)b - (b�a)c = [a₂b₁c₂-a₂b₂c₁+a₃b₁c₃-a₃b₃c₁, a₁b₂c₁-a₁b₁c₂+a₃b₂c₃-a₃b₃c₂, a₁b₃c₁-a₁b₁c₃+a₂b₃c₂-a₂b₂c₃]

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 b-c 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 b-c plane.  So the vector that results from a�(b�c) is in the b-c 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:

1. The scalar factor for b is the dot product of the other two vectors, and likewise for c.
2. 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.
3. Use the right-hand 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.

Jacobi Identity

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 = [a₁,a₂,a₃], and let b and c be defined similarly.

a�b = [a₂b₃-a₃b₂, a₃b₁-a₁b₃, a₁b₂-a₂b₁]

(a�b)�c = [(a₃b₁-a₁b₃)c₃-(a₁b₂-a₂b₁)c₂, (a₁b₂-a₂b₁)c₁-(a₂b₃-a₃b₂)c₃, (a₂b₃-a₃b₂)c₂-(a₃b₁-a₁b₃)c₁]
(a�b)�c = [a₃b₁c₃-a₁b₃c₃-a₁b₂c₂+a₂b₁c₂, a₁b₂c₁-a₂b₁c₁-a₂b₃c₃+a₃b₂c₃, a₂b₃c₂-a₃b₂c₂-a₃b₁c₁+a₁b₃c₁]

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 b-c plane
(a�b)�c = (c�a)b - (b�c)a   -- because the resulting vector is in the a-b plane

Call 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

 

Area of Triangle formed by Endpoints of Three Vectors

Three Vectors a, b, and c have a common initial point. Their endpoints form a triangle whose area is given by the magnitude of the vector:

1/2(a�b + b�c + c�a)

Proof

Triple Product of Cross Products

(a�b)�((b�c)�(c�a)) = (a�(b�c))²

Proof

Internet references

Related Pages in this website

Vector Dot Product

Vector Cross Product

Triple Product -- a·(b�c) is a scalar representing the "signed volume" of a parallelepiped

Triangle Area Using Vectors, part 1 and part 2

Triangle Area using Determinant

Matrix Math

Geometry and Trigonometry, and in particular, the Points and Lines section.

The webmaster and author of this Math Help site is Graeme McRae.