(13) Finding equations of planes using normal vectors
The direction perpendicular to a plane is unique to that plane (and
any plane parallel to it).
To see how we can use this to give us another form of the vector
equation of a plane, we'll start with the case where the plane
passes through the origin. (It's particularly easy to see how to do
it in this case!)
I've shown part of such a plane in my drawing below.
Suppose we know the vector N which is perpendicular to the plane.
This means that it must be perpendicular to the position vector r
of any
point in the plane from the origin, so
the dot product of the perpendicular
vectors N and r
gives us the equation
N.r = 0.
The vector N is called a normal
vector to the plane.
(Any vector
parallel to the N I have drawn will also be a normal vector
to this plane and will work equally well.)
Now we extend this method to find the equation of a plane which
doesn't pass through the origin.
I've shown part of such a plane in the drawing below.
This time, we have to be able to get to the plane first from the
origin, so we must know the position vector of some particular point in the plane from the
origin.
In my drawing, this point is M with position vector m.
If P is any general point in the plane, so that the
vector MP = p, then N and p are perpendicular to
each other.
Therefore N.p = 0 but p = r - m.
N.(r - m) = 0 or N.r = N.m = C
where C is the number we get from working out the dot product of the
two known vectors N and m.
Now suppose we are starting with the plane below, with the 2 known
vectors s and t lying in this plane.
How could we use s and t to find a vector N
which would be perpendicular to this plane?
It's worth thinking about this before finding
out.
Or return to the previous section,
or return to the vectors homepage.