Bob

ps

I made it

which shows what I mean about many answers being possible.

]]>And I think we have it, thank you so much

]]>I see what's wrong. You cannot use

because that's a position vector for the original plane. You need one for the new plane so choose from

Then it will work. :)

Bob

]]>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

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

Now if I can choose any value for *a* then I may as well go with 1, which gets us:

Okay so let's choose:

Now, at this point the most sensible thing to do would seem to be to multiply *n* by 10, but I happen to know that the book's answer actually requires me to multiply by -10. Now, unless I'm very much mistaken, the values of *n* make no difference at all, but it'd be nice to get the book's value of *n*, just to tie it off nicely. So we have:

Which leaves me with something of a problem, because the correct answer should be 120

]]>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

]]>So, firstly we begin by picking three points, presumably from

So we could say let

Which of course gives

Respectively, unless I'm very much mistaken. Now, if I have understood correctly so far we must now transform these points with **T**, which, I believe, gives:

But at this point I don't fully understand AB = AO - OB etc.

]]>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

]]>temp = 3 celcius

B

I see you are logged on at the moment. I've just come in from gardening to have lunch. I'll keep the computer running but you'll have to wait until later for the example as I want to take advantage of the light and fine weather to complete the garden jobs.

Bob

]]>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.

]]>Finally got it sorted. The method outlined in post 26 is sound. When I worked out a, c and b I got them in that order,

but failed to write them correctly for the vector 'n'.

The book method seems a bit complicated to me as you land up with such a tricky matrix multiplication to do.

Given what I'm like with little errors I'd never get it right that way. But if you want to try it, it'll probably be good for you.

To get the equation for the second plane in the required format you have got to find a way to compute 'n',

the vector that is perpendicular to the plane.

My method gets you two vectors in the plane, and it is fairly straight forward to get 'n' from them.

Then you can use any point in the plane to find 'p'.

If you are still finding this tough I have a 2D example I can send. I made it up when I was getting desperate to sort out why my results were inconsistent.

I wanted to check I had the theory soundly worked out.

And the answer to my question

When T transforms a plane it (i) transforms position vectors for points in the plane OR (ii) it transforms vectors that lie in the plane.

is definitely (i), which was what I had worked with all along.

Let me know if you have managed to complete the equation of the plane or want more help.

Best wishes,

Bob

ps. When this thread moved onto page 2 (at post 26), the page width went wrong and had stayed so for me ever since. How does it display for you?

]]>Did you mean that? You cannot make a plane with the same vector for all three components.

Bob

]]>