### (16) Equations of planes in the form Ax + By + Cz = D

This is another useful way to describe planes. It is known as the cartesian form of the equation of a plane because it is in terms of the cartesian coordinates x, y and z.
The working below follows on from the pages in this section on finding vector equations of planes and equations of planes using normal vectors.
The form Ax + By + Cz = D is particularly useful because we can arrange things so that D gives the perpendicular distance from the origin to the plane.
To get this nice result, we need to work with the unit normal vector. This is the vector of unit length which is normal to the surface of the plane. (There are two choices here, depending on which direction you choose, but one is just minus the other).
I'll call this unit normal vector n.
Next we see how using n will give us D, the perpendicular distance from the origin to the plane.
In the picture below, P is any point in the plane. It has position vector r from the origin O.
Now we work out the dot product of r and n.
This gives us r.n = |r||n|cos A.
But |n| = 1 so we have r.n = |r|cos A = D.
This will be true wherever P lies in the plane.

Next, we split both r and n into their components.
We write r = xi + yj + zk and n = n1i + n2j + n3k.
Therefore r.n = (xi + yj + zk) . (n1i + n2j + n3k) = D
so r.n = xn1 + yn2 + zn3 = D.
We see that n1, n2 and n3 (the components of the unit surface normal vector) give us the A, B and C in the equation Ax + By + Cz = D.

### A numerical example

I've put this in here so that you can see everything actually happening and see how it ties back to the earlier pages in this section.

We'll let s = i - 6j + 2k and t = 2i - 2j - k

We'll take m, the position vector of the known point M in the plane, to be
m = 2i + 3j + 5k.
P is any point in the plane, with OP = r = xi + yj + zk.

First, we find N, a normal vector to the plane, by working out the cross product of s and t.

This gives s x t = 10i + 5j + 10k = N.
The length of this vector is given by the square root of (102 + 52 + 102) = 15.
So the unit normal vector, n, is given by
n = 1/15(10i + 5j + 10k) = 2/3i +1/3j + 2/3k.

Now we use n.r = n.m = D to get the equation of the plane.
This gives us
(2/3i +1/3j + 2/3k).(xi + yj + zk) = (2/3i +1/3j + 2/3k).(2i + 3j + 5k)
or 2/3x + 1/3y + 2/3z = 4/3 + 3/3 + 10/3 = 17/3.
The perpendicular distance of this plane from the origin is 17/3 units.

So what would have happened if we had found the equation of the plane using the first normal vector we found?
Using N.r = N.m gives
(10i + 5j + 10k).(xi + yj + zk) = (10i + 5j + 10k).(2i + 3j + 5k)
or 10x + 5y + 10z = 20 + 15 + 50 = 85.
It is exactly the same equation as the one we found above except that it is multiplied through by a factor of 15, and 85 gives us 15 times the perpendicular distance of the origin from the plane.

Also, are you confident that you will get the same equation for the plane if you start out with the position vector of a different known point in it?
The point L also lies in this plane. Its position vector l is given by l = 7i - 7j + 5k.
Check that working with l instead of m does give you the same equation for the plane.
Geometrically, you can see that this will be so.

L and M are both just possible positions of P, so that both n.l and n.m give the distance D.
Try one for yourself!
The two vectors s = 4i + 3k and t = 8i - j + 3k lie in plane Q.
The point M also lies in Q and its position vector from the origin is given by
m = 2i + 4j + 7k.

Show that the perpendicular distance of the origin to this plane is 2 units and find its equation.

### The general case

This is how the working goes with letters taking the place of the numbers we have used in the numerical example.
m is the position vector of the known point in the plane.
n is the unit surface normal to the plane.
We'll let m = x0i + y0j + z0k and n = Ai + Bj + Ck.
The position vector of the general point P in the plane is given by
r = xi + yj + zk where the values of x, y and z vary according to the particular P chosen.

Now we use n.r = n.m = D to write down the equation of the plane. This gives us

(Ai + Bj + Ck) . (xi + yj + zk)= (Ai + Bj + Ck) . (x0i + y0j + z0k). = D
so Ax + By + Cz = Ax0 + By0 + Cz0 = D
or, if you prefer, you can write
A(x-x0) + B(y-y0) + A(z-z0) = 0.

If you have found a normal vector which is not of unit length, you will first need to scale it down.
Suppose you have found N = N1i + N2j + N3k.
Then the length of N is given by

and n, the unit normal vector, is given by
Now, putting n = Ai + Bj + Ck, we have

The following two pages look at how we write the equations of lines and planes in 3 dimensions in terms of x, y and z.

What does the equation 3y + 4x = 12 give us?

or back to the previous page