(8) The vector or cross product
Is it possible to find a way to multiply two vectors together so that we
get a vector result?
The problem is that, working from
the two directions of our starting vectors, we have to find
some way of using these to give us the single direction of the
This problem makes it impossible to define this kind of vector
multiplication in 2 dimensions but there is a very neat way of
solving it in 3 dimensions.
We first slide the two vectors together, so
that their tails meet. (As long as at least one of them is a free vector,
we will always be able to do this.)
Now they both lie in one particular flat surface
or plane. We can use the direction perpendicular to this plane to
give us the direction of our vector product. We only have to decide which
of the two possible perpendicular directions to choose.
The drawing below shows how we do this to find the vector product of
the two vectors a and b.
I've redrawn b shifted so that the tails of a and b
We now define the vector product of a and b as
a x b = |a||b| sin t n
where n is a unit vector perpendicular to the plane in which
a and b lie. (I've shown n as part of the
vector a x b.)
We choose the direction of n so that
a, b and n can fit along the thumb, first finger
finger of your right hand. (Check this with your hand and my drawing.)
Notice that this means that the direction of b x a is given
by - n
so a x b = - b x a.
The multiplication sign used to show the vector product is called 'cross'.
It is also sometimes written as a little upside down v like a
(If you are visiting from finding a
normal vector, you can return here.)
This definition has some very neat practical applications.
The drawing below shows one of these.
If we have a force F acting through a
point P with position vector r with respect to O,
then F and r lie in
a plane through O. I have also redrawn the force vector F
shifted so that its tail is at O.
The torque or
moment of F about
an axis through O perpendicular to this plane is given by
T = r x F = |r||f| sin t n
This torque measures the turning effect of F about the axis
through O. It is independent of the
position of P on the line of action of F.
You can see that this must be so from this drawing looking down at the
plane containing the line of action of F and the origin, showing the
position vectors of 3 different choices of point on this line of action.
We have |p| sin P = |q| = |r| sin R so the torque
of F about the perpendicular axis through O is independent of the
point you choose.
My next drawing shows that the value of sin t does indeed give the various
practical possibilities, when a person applies torque using
a lever to turn on a tap.
Here's another physical application of the cross product.
In the drawing below, I've shown a particle of mass m with position
vector r relative to
an origin O. If this particle has
velocity v then its
momentum p is given by p = mv.
The angular momentum or
moment of momentum
of the particle is defined as
angular momentum = r x p = r x mv
= m(r x v).
So not only does this definition of vector multiplication make sense
mathematically but, as I've shown, it also has extremely useful
Some special cases
What answers do we get if we work out the cross products for the different
possible pairs we can choose from the unit vectors i, j and
Since sin 0 = 0 and sin 90 = 1 and each vector is of unit length,
i x i = j x j = k x k = O,
(the zero vector).
Also, i x j = k and j x k = i
and k x i = j
while j x i = - k and
k x j = - i and
i x k = - j.
You can see how the various
plus and minus signs come by using the right-hand rule for each
I've shown this for the case i x j = k.
Now we can see
how vector multiplication will work out using components.
Suppose we have
a = a1 i + a2 j + a3
and b = b1 i + b2 j
+ b3 k.
Then a x b =
(a1 i + a2 j + a3 k)
x (b1 i + b2 j + b3 k).
It can be shown that it is all right to work this out by working out all
the 9 separate little cross products that we get from multiplying these
two brackets together. Doing this, and using the results above, we get
a x b =
(a2b3 - a3b2) i +
(a3b1 - a1b3) j +
(a1b2 - a2b1) k
and this gives us the rule for how we work out a cross product using
Here is a numerical example using this result.
A force F = 3i + 2j + 4k acts through the
point with position vector r = 2i + j + 3k.
What is its torque about a perpendicular axis through O?
The torque = r x F = (1x4 - 3x2) i + (3x3 - 2x4) j +
(2x2 - 1x3) k
= - 2i + j + k.
This method is very much easier than trying to work out the moment of a
force in 3 dimensions using geometry.
(If you are visiting from can we multiply 3 vectors
together? or finding a normal vector, you
can use these links to return.)
Can we also multiply vectors in threes?
(We'll need to know which two we should start with.)
Which of the following will work?
Two are impossible. One gives a vector answer and one gives the volume of
a neat little box.
- (2i . 3i) . 4j
- (2j . 4j) x 3k
- (2i x 3j) x 4i
- (2i x 3j) . 4k
Which is which? How big is the box? It really is worth working out the
answers to these for yourself before continuing.
or back to the previous section
or back to the vectors homepage.