HANDOUT ONE: MORE ON DOT PRODUCTS AND CROSS

PRODUCTS

PETE L. CLARK

Few things are more basic to the study of geometry in two and three dimensions
than the dot and cross product of vectors. On the philosophy that it is good to
understand simple things very well, I give here a treatment of these two products
which is more complete than the one in the textbook (and than the presentation
given in class). The emphasis is on an understanding of the following two product
formulas:

For any vectors v, w in R2 or in R3 , if the angle between them is θ, then

(1) v · w = ||v|| ||w|| cos θ.
3
For any vectors v and w in R (only!), if the angle between them is θ then
(2) ||v × w|| = ||v|| ||w|| sin θ.
Moreover, the direction of the cross-product is determined by the following infor-
mation: v × w is perpendicular to both v and w, and the triple (v, w, v × w) is
oriented according to the right-hand rule.

Your textbook takes these equations as the definitions for the dot and cross product
of two vectors. I prefer to regard them as properties of the operations, which are
defined directly by algebraic formulas.

1. The dot product formula

Let us look first at the dot product: if v = (a, b, c) and w = (x, y, z) then
v · w = (a, b, c) · (x, y, z) := ax + by + cz.1 The dot product of vectors in the
plane is defined similarly, just without the last coordinate: (a, b) · (x, y) := ax + by.
Indeed, the dot product can be defined for pairs of vectors with n components as
(x1 , . . . , xn ) · (y1 , . . . , yn ) := x1 y1 + x2 y2 + . . . + xn yn .

Whenever you see a new operation – especially a new kind of “multiplication”

– you should not assume that it has all the nice properties that multiplication of
ordinary numbers. Notice that the dot product is a rather weird thing: given two
vectors, we combine them to get a scalar, which may not remind us of anything at
all. Here are some properties:
v·w =w·v
u + (v · w) = (u · v) + (u · w)
(cv) · w = v · (cw) = c(v · w)

1The “:=” is computer science notation meaning that the quantity to the left of the symbol is
defined to be equal to the quantity to the right of the symbol.
1
2 PETE L. CLARK

The last one says that if we scale either one of the vectors in the dot product, then
the dot product scales by the same factor. Thus
v · w = (||v||v̂) · (||w||ŵ) = ||v|| ||w||v̂ · ŵ.
Recall here that the notation v̂ for a unit vector in the direction of v was suggested
(and used!) in Wednesday’s class. Thus for every vector v (except the zero vector)
we have v = ||v||v̂ expressing v as a magnitude and a direction. Thus to prove the
dot product formula, it suffices to show that the dot product of two unit vectors is
the cosine of the angle between them.

This is easy for vectors in R2 : to say they are unit vectors just means that they lie
on the unit circle. Thus v̂ = (cos α, sin α) and ŵ = (cos β, sin β) say, where v̂ make
angles of α and β with the x-axis respectively. Note that the angle between them
is θ = β − α (draw a picture!). Now we compute
v̂ · ŵ = (cos α, sin α) · (cos β, sin β) = cos α cos β + sin α sin β =
cos(α − β) = cos(β − α) = cos(θ).
In three dimensional space, we had better do something more conceptual lest we get
embroiled in spherical geometry. The dot product of two unit vectors is supposed
to depend only upon their relative position, so if we rotate both vectors the same
amount about any axis, their dot product is supposed to be unchanged. (Even
so, the derivation is more involved than I had thought, and uses some ideas on
parameterized curves that we will discuss next week.)

Suppose P = (x, y, z) and we rotate the vector OP about the x-axis. (Stop
and try to picture this.) The point describes a circle, perpendicular to the x-
axis, whose radius is the p distance from P to the x-axis, which is ||P − projî P || =
||(x, y, z) − (x, 0, 0)|| = y 2 + z 2 = R, say; so the equations describing it are
(x, R cos θ, R sin θ); we get the point P itself when tan θ = z/y. Now take two
vectors v and w and represent them in this way:
v = (x1 , R1 cos θ1 , R1 sin θ1 )
w = (x2 , R2 cos θ2 , R2 sin θ2 )
We have
v · w = x1 x2 + R1 R2 (cos θ1 cos θ2 + sin θ1 sin θ2 ) = x1 x2 + R1 R2 cos(θ1 − θ2 ).
Now rotate both vectors through any common angle α about the x-axis, getting
vα = (x1 , R1 cos(θ1 + α), R1 sin(θ1 + α)), wα = (x2 , R2 cos(θ2 + α), R2 sin(θ2 + α)).
Now we compute
vα · wα = x1 x2 + R1 R2 (cos(θ1 + α) cos(θ2 + α) + sin(θ2 + α) sin(θ2 + α)) =
x1 x2 + R1 R2 (cos((θ1 + α) − (θ2 + α)) =
x1 x2 + R1 R2 cos(θ1 − θ2 ) = v · w.
So indeed, when we rotate a pair of vectors about the x-axis, their dot product
doesn’t change. By symmetry, the same holds for rotation about the y-axis and the
z-axis2 But now consider that starting with any pair of vectors in R3 , by rotating
2In other words, we see that if we wrote the coordinates in a different order, the calculation
would be essentially the same.
 HANDOUT ONE: MORE ON DOT PRODUCTS AND CROSS PRODUCTS 3

them about the three coordinate axes we can put both vectors in the xy-plane –
since we moved them jointly, the angle between them is unchanged. Finally we can
apply the dot product formula in R2 .

An important consequence of the dot product formula (and the derivation!) is

that it shows that the dot product of two vectors is intrinsic: if we took two
wooden sticks of length ||v|| and ||w|| respectively and attached them to each other
with a fixed angle θ between them and pinned the vertex to the origin in R3 , then
no matter how we spin this contraption about, the dot product remains the same!

2. The cross product formula

It cannot be overemphasized that the dot product of two vectors is not really a
“product” at all in the sense that it given two objects of the same sort (vectors), it
returns an object of a different sort (a scalar, or number). It seems more natural
to look for vector products of two vectors. Indeed, one might be tempted to look
at the following operation:
v ? w := (a, b, c) ? (x, y, z) = (ax, by, cz)
In other words, we try to multiply the vectors the same way we add them: compo-
nentwise. This operation is perfectly well-defined; the problem is that, unlike the
dot product and the cross product to be seen shortly, it has no use, or more pre-
cisely, no geometric or physical interpretation. For instance, the dot product is zero
if and only if the vectors are perpendicular to each other, whereas v ? w = (0, 0, 0)
only if either a = 0 or x = 0, either b = 0 or y = 0 and either c = 0 or z = 0.
There’s nothing here worthy of our attention.

On the other hand, there is a much crazier looking product, which turns out to
be far more useful, namely the cross product:
v × w = (a, b, c) × (x, y, z) := (bz − cy, −(az − cx), ay − bx).
To remember the formula, we can give it in a more succinct way, using determinants:
let î = (1, 0, 0), ĵ = (0, 1, 0), k̂ = (0, 0, 1) be the unit vectors along the x, y and z
axes respectively. Then the cross product of v and w can be computed as a 3 × 3
determinant of the “matrix”3  
î ĵ k̂
 a b c 
x y z
or î(bz − cy) − ĵ(az − cx) + k̂(ay − bx).

The most important property of the cross product of v and w is that it is per-
pendicular to both v and w. Let’s check this:
(v×w)·v = (bz −cy, cx−az, ay−bx)·(a, b, c) = abz −bcy+bcx−abz +acy −bcx = 0.

(v×w)·w = (bz−cy, cx−az, ay−bx)·(x, y, z) = bxz−cxy+cxt−ayz+ayz−bxz = 0.

3The quotation marks are there because some of the entries in the matrix are themselves
vectors and others are scalars, which is not the usual state of affairs.
4 PETE L. CLARK

We would also like to know the magnitude of the cross product: recall that we are
supposed to have the formula
||v × w|| = ||v||||w|| sin θ.
This is a very exciting thing for the magnitude to be: it implies that the cross
product is only zero when the angle between v and w is zero, i.e., unless v and w
are scalar multiples of each other. Moreover, the parallelogram formed by adding
v and w has base length ||w|| and height ||v|| sin θ, so the magnitude of the cross
product is precisely the area of this parallelogram. Finally, we cannot help but be
struck by the similarity to the formula for |v · w|, which is the same except with
cos θ. This allows us to guess the Lagrange identity
||v × w||2 + |v · w|2 = ||v||2 ||w||2 .
Now we are in a sneaky situation: we can apply “Littlewood’s Principle” which is
(no kidding): purely algebraic identities are very easy to check once someone else
has written them down for you. (In other words, geometry and trigonometry can
get difficult, but purely algebraic manipulations are always rather routine.) So:
||(a, b, c)×(x, y, z)||2 +|(a, b, c)·(x, y, z)|2 = (bz−cy)2 +(az−cx)2 +(ay−bx)2 +(ax+by+cz)2 =
b2 z 2 −2bcyz+c2 y 2 +a2 z 2 −2acxz+c2 x2 +a2 y 2 −2abxy+b2 x2 +a2 x2 +b2 y 2 +c2 z 2 +2abxy+2acxz+2bcyz =
(a2 + b2 + c2 )(x2 + y 2 + z 2 ) = ||v||2 ||w||2 .
Using the Lagrange identity and the dot product formula, we can derive the formula
for the magnitude of the cross product, namely:
||v × w||2 = ||v||2 ||w||2 − |v · w|2 =
||v||2 ||w||2 (1 − cos2 θ) = ||v||2 ||w||2 sin2 θ
and taking square roots we get ||v × w|| = ||v||||w|| sin θ.

The final bit about the cross product is its orientation: we know it is perpen-
dicular to v and to w and has a certain magnitude, so this determines it up to a
sign. Why does it come out to be the right-hand rule – and indeed, how could we
see such a thing without contorting our poor right hand to every possible angle? It
comes down to the simple fact that
î × ĵ = k̂
as you should check directly, so the right-hand rule holds in this one case. Now,
the cross product was given as a simple polynomial function in the entries of the
two vectors, so it is a continuous function of v and w: if we change v and w just
a little bit, so too is the change in v × w small. So as we start with î × ĵ and slowly
rotate our hand to put the fingers at v and the ball of the hand at w, it would be
a big change if at any given time the cross product suddenly flipped to be in the
opposite direction than the right hand rule predicts. In other words, given that the
right hand rule holds for some pair of vectors, a failure of it to hold at any other
pair would mean a discontinuity in the cross product, which is not possible.