Coordinate Formula For Cross Product
Learning goals: use the ideas we’ve developed to discover the coordinate formula for cross
products.
Now we know the triple product and how to compute it. Let x = (x1, x2, x3), y = (y1, y2, y3), and z
= (z1, z2, z3). The triple product is
x1
x2
x3
y1
y2
y3 . We also know that this is x⋅(y × z).
z1
z2
z3
Now we would like to find a formula for y × z = (a, b, c). How can we find a, b, and c? Well,
another way to think about it is that y × z = ai + bj + ck. Now we know how to find a, b, and
c—since the i, j, and k are mutually orthogonal. So we only need projections. What’s that you
say? They’re actually orthonormal? So we really only need dot products. That is a = (y × z)⋅i,
which we will write as i⋅(y × z) =
1
y1
0
y2
0
y3 . We can easily compute this either using
z1
z2
z3
reduction or our diagonals trick: a = y2z3 – y3z2. We can find similar formulas for b and c. So we
now have a coordinate formula for cross products:
y × z = (y2z3 – y3z2)i + (y3z1 – y1z3)j + (y1z2 – y2z1)k
Now this is not a very easy formula to remember, so we seek something a little nicer. The idea is
to look at the formula above and compare it to the determinant formula. In that formula, we also
had terms with y2z3 and y3z2 in them—and both were multiplied by x1. And so forth.
Substituting i for x1 and so forth, we get the nifty formula y × z =
î
ĵ
k̂
y1
y2
y3 .
z1
z2
z3
Now as we were looking at the algebra of the cross product, we temporarily side-stepped
distributivity. It works, but is hard to prove geometrically (though it can be done—when trying
to figure out x × (y + z) we first start by breaking y and z into parts parallel to x and parts
perpendicular to x, and the homework will show that we can ignore the parallel parts; the
perpendicular parts are at least easier to work with since now all the angles are right angles).
We could use our coordinate formula, though that would also be kind of messy. So we’re going
to use a nifty trick based on the ability to interchange dot and cross products in the triple product.
First, a lemma:
Lemma: x = y if and only if x⋅a = y⋅a for every vector a.
Proof: certainly if x = y then x⋅a = y⋅a for any choice of a. Conversely, let x⋅a = y⋅a for every
vector a. Choose a = x – y. Then x⋅(x – y) = y⋅(x – y), or x⋅(x – y) - y⋅(x – y) = 0, which we can
simplify to (x – y)⋅(x – y) = 0. So ||x – y|| = 0, so x = y.
Theorem: x × (y + z) = x × y + x × z.
Proof: Let a be any vector, and consider the triple product a ⋅ ( x × (y + z) ) . Because we can
swap the dot and the cross, this is equal to (a × x)⋅(y + z). Now we know that the dot product is
distributive, so this becomes (a × x)⋅y + (a × x)⋅z = a⋅(x × y) + a⋅(x × z) = a⋅(x × y + x × z). So
the two are equal when dotted with any a, so by the lemma the two vectors are equal.
We could have used this to find our coordinate formula for cross products, as well:
(y1, y2, y3) × (z1, z2, z3) = (y1i + y2j + y3k) × (z1i + z2j + z3k). Now distribute this: y1z1(i × i) +
y1z2(i × j) + !. Terms like i × i are zero, while we already know i × j = k, etc.