Hmmmm it's a tricky one, I copied out the question exactly as it stands in the book, but if it helps I can copy out the book's example of a problem which is similar to these questions, but not quite the same, perhaps that would shed some light:

Last edited by Au101 (2011-01-03 00:36:24)

Hi Au101,

Thanks for that. 

I'm often making numeic errors, so I'm going to create Excel formulas for the planes and transformations so I don't keep slipping up.

Then I can sub in any plane and any matrix and quickly generate vectors.  I'll use the book example to check my formulas.

Then I'll come back to you.

Looking at that example I think it has answered my question, anyway.

Bob

Last edited by bob bundy (2011-01-02 04:51:04)

Oh, whoops, sorry, no I didn't I copied and pasted the matrix template for simplicity and forgot to change the entries! Sorry for the confusion, it should be the same as the last line of the preceding post, I'll just change it.

Ooooh thanks ever so much for all of your tribulations Bob. I suppose that either method would suit me fine, I'm just not entirely sure I'm clear on your method, I don't suppose you could provide an example, or further instruction to clarify? (I'm sorry to request further tribulations). Otherwise the book method is perfectly fine, but I don't really see how to move from the book's finding the vector equation to finding an equation in the form

Oh and yes I'd meant to say, the screen is much wider than normal, it may have been to do with my using the LaTeX text environment, which does not drop lines automatically. I tried to fix it, but it doesn't seem to have made much of a difference I'm afraid.

Last edited by Au101 (2011-01-03 00:53:34)

Of course, I'm perfectly happy to wait, especially for answers as instructive as yours. However, I'm surprised to here about the weather, it's very cloudy here on (presumably) the other side of Essex.

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 form

and
form 3

The 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

Last edited by bob bundy (2011-01-03 05:07:51)

hi

Ok so far.

If we give your points letters,  (don't forget, they are position vectors with respect to the origin, O)

then

Now do that again for BC = BO + OC

Can you continue from here?

Bob

Last edited by bob bundy (2011-01-03 07:53:09)

hi Au101,

Interesting.  I got 120.  It'll take a while to find the differences and I'm in the middle of de-salting my car (again while the weather holds).  I'll come back to this when the sun goes down.

Bob

Ah, okay, so let's say

And I think we have it, thank you so much

Last edited by Au101 (2011-01-04 03:16:24)