hi Au101

OK. Here we go.

There are three forms for the equation of a plane (that I can think of, there may be more).

They are:

form 1

form 2; the dot product formand

form 3The first form is derived from the diagram in post 20.

This is the position vector for a point in the plane

and

are two vectors lying in the plane.

In the second form

is a vector that is perpendicular to all vectors lying in the plane. r and n are being 'dotted' together and the result is a constant.

If you then replace 'r' by

and do the dot product

you get the cartesian version that is the third form.

From this you can see that form 2 and form 3 are the same plane.

Less obviously, so is form 1.

Proof:

In form 1 put lambda = 1 and mu = 1 and you get x = 3, y = 3, z = 1

put lambda = 1 and mu = -1 and you get x = 1, y = -1, z = -1

put lambda = 0 and mu = 1 and you get x = 2, y = 3, z = 2

If you try each of these sets of values in x - y + z you get 1 every time.

So form 1 and form 3 have three non-collinear points in common. That's enough to prove they represent the same plane because you only need three points to define a plane (provided they are not in the same straight line).

Now to the question.

You know the equation of the plane before the transformation. The book method transforms points directly from this equation to get the equation after the transform.

I think the algebra for this is a bit horrid and

you still have to convert it into form 2.

So my method was to pick three points in the plane and transform them. Let's call them A, B and C. That's just number work. ***

Now get two vectors in the plane by doing AB = AO - OB etc.

Any two will do because, provided they are not parallel, any two vectors will 'span' the plane ie. will enable you to reach all points in the plane.

Now to get the vector that's at right angles to both these vectors.

Imagine by some trick of gravity you can stand on the plane with 'up' meaning 'at right angles to' the plane. If I asked you to point a stick at right angles to the plane it would go straight up. If you draw any line on the 'ground' = 'the plane' , it would be at right angles to the stick. And it wouldn't matter how long the stick was. A stick twice as long, would still be at right angles to every line in the plane.

Call the vector at right angles

So do a dot product between one vector in the plane and 'n' and set it equal to zero. Do again for the second vector.

You've got 2 equations with a, b and c as unknowns. Choose any 'a' to make the calculations easy. That will enable you to work out 'b' and 'c'. That gives you a possible 'n' the vector perpendicular to the plane. You might think it is cheating to choose 'a'. But if you had chosen an 'a' that was twice as big, you'd have got 'b' and 'c' twice as big as before so all that would happen is you'd get '2n' for the perpendicular vector. You can use

any vector that is at right angles so the first would do!

Now to get 'p'.

The equation

is the equation for all points, 'r', in the plane.

Back at point *** we had three possible points so 'sub' in any one set of x, y and z values and you'll get 'p'

Check by 'subbing' in the other values from *** to see if you get the same p.

Problem done!

Does that all make sense?

Bob