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:
- 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 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.
