(7) Finding the direction cosines of a vector
Suppose we have a point P with position vector p and that this vector
makes angles of A, B and C with the x, y and z axes respectively.
I've drawn this below.
We'll start by finding the angle which p makes with the x axis.
Suppose p = ai + bj + ck written in component
form.
Then p.i = |p||i| cos A using section (6).
But p.i = (ai + bj + ck).i = a.
Also, |i| = 1 because i is a vector of one unit in length.
So now we have
a = |p| cos A and therefore cos A = a/|p|.
Similarly, cos B = b/|p| and cos C = c/|p|,
and from section (4) we know that |p| = the square root
of (a2 + b2 + c2).
We have now found the cosines of the angles which p makes with the
x, y and z axes respectively. These are called
direction cosines.
Now the vector (cos A)i + (cos B)j + (cos C)k is rather
special.
Since it is equal
to (a/|p|)i + (b/|p|)j + (c/|p|)k
it has the same direction as p and lies along OP.
Also, its length is given by
the square root of
(cos2A + cos2B + cos2C).
From above, this is the same as
the square root of {(a2 + b2 + c2)
divided by (a2 + b2 + c2)} = 1.
So it is a unit vector in the direction of p,
and it is often written by putting a little
circumflex hat over the p and calling it p hat.
If we had started with a free vector p, the working to find its
direction cosines would be exactly the same, since the angles it makes
with the 3 axes remain the same if we slide it until its tail is at the
origin.
We saw in Section (6) that defining the multiplication of two vectors
as a scalar product gives us the useful practical application of finding
work done by a force. In the next section we discover
how it is possible to define the multiplication of two vectors in an
alternative new way so that the answer is also a vector. This definition also has
extremely useful practical applications.
Next
or back to the previous section
or return to the vector homepage.