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

I'll call this unit normal vector

Next we see how using

In the picture below, P is any point in the plane. It has position vector

But

Next, we split both **r** and **n** into their components.

We write **r** = x**i** + y**j** + z**k** and
**n** = n_{1}**i** +
n_{2}**j** + n_{3}**k**.

Therefore
**r.n** = (x**i** + y**j** + z**k**) **.**
(n_{1}**i**
+ n_{2}**j** + n_{3}**k**) = D

so **r.n** = xn_{1} + yn_{2} + zn_{3} = D.

We see that n_{1}, n_{2} and n_{3}
(the components of the unit surface normal vector) give us
the A, B and C in the equation Ax + By + Cz = D.

We start with the plane I show below.

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

P is any point in the plane, with OP =

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

So the unit normal vector,

Now we use

This gives us

(2/3

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

(10**i** + 5**j** + 10**k**)**.**(x**i**
+ y**j** + z**k**) =
(10**i** + 5**j** + 10**k**)**.**(2**i** + 3**j** + 5**k**)

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** = 7**i** - 7**j** + 5**k**.

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.

Try one for yourself!

The two vectors

The point M also lies in Q and its position vector from the origin is given by

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

We'll let

The position vector of the general point P in the plane is given by

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

so Ax + By + Cz = Ax

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** =
N_{1}**i** + N_{2}**j** + N_{3}**k**.

Then the length of **N** is given by

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.

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