Vector Equations

Bob · 2011-01-01 03:08:46

Hi Au101

I thought this was straight forward but I'm now on explanation #3 and I still cannot get it right.

This is what ought to work.

Choose three sets of values for 's' and 't' so that you have the position vectors for three points in the plane. **

Transform them using T to get three position vectors for points in the new plane. (say A, B and C)

Find vectors that lie in the new plane; AB and BC would do.

Let the vector perpendicular to the new plane be

and make equations with a, b, and c by dot products AB.n = 0 and BC.n = 0

That gives two equations and three unknowns; not enough to find a single 'n',
but any multiple of 'n' will also work so choose a value for, say, 'a'
and then use the equations to find b and c. That will give you a solution for 'n'.

(If you want, you could try another value of 'a' and generate new values for b and c.
Although different, you will find this second solution is just a multiple of the first so either will do.)

So now you know 'n'.

You know that the transform of

is position vector for a point in the new plane so substitute into r.n = p to find the constant 'p'.

I've done all that and got an equation for the new plane.

Now to check it. The other position vectors should also fit this solution.

So far I haven't managed to acheive this!

So, I think my calculations have an error but I cannot find it. I don't think the logic is faulty (I hope).

Do you want to try this and post your figures. Then I can compare them with mine.

Bob

ps. Explanation #1 had a different line at the point marked **

Instead of transforming three position vectors, I tried transforming the two vectors
lying in the first plane and assuming the result was two vectors in the second plane.
That gave a completely different solution and I don't think it is correct.

Does the book make the following point clear:

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.

Last edited by Bob (2011-01-01 22:30:45)

Au101 · 2011-01-01 04:49:15

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 (2011-01-02 01:36:24)

Au101 · 2011-01-01 04:51:25

I don't know if that will help but that's the book's angle, I can't find any further clarification in the book I'm afraid. I'd be happy to have a go at getting my own figures, except I'm not entirely sure I understand what to do.

Last edited by Au101 (2011-01-01 06:30:48)

Bob · 2011-01-01 05:47:24

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 (2011-01-01 05:51:04)

Bob · 2011-01-01 06:17:15

Your last post:

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

Bob

Au101 · 2011-01-01 06:30:20

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.

Bob · 2011-01-01 22:41:23

hi Au101,

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?

Last edited by Bob (2011-01-01 22:46:15)

Au101 · 2011-01-02 01:49:17

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 (2011-01-02 01:53:34)

Bob · 2011-01-02 01:53:30

hi

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

Au101 · 2011-01-02 01:55:25

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.

Bob · 2011-01-02 02:03:43

I've got white fluffy clouds and some blue bits.

temp = 3 celcius
B

Last edited by Bob (2011-01-02 02:04:28)

Bob · 2011-01-02 06:05:36

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 (2011-01-02 06:07:51)

Au101 · 2011-01-02 07:35:10

Wow what an explanation thank you so much.

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.

Last edited by Au101 (2011-01-02 07:36:06)

Bob · 2011-01-02 08:52:15

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 (2011-01-02 08:53:09)

Au101 · 2011-01-03 02:06:35

Well, we only need one other vector, so we may as well do BC

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

Bob · 2011-01-03 02:34:48

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

Bob · 2011-01-03 04:02:24

Hi again,

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

Au101 · 2011-01-03 04:14:16

Ah, okay, so let's say

And I think we have it, thank you so much

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

Bob · 2011-01-03 04:48:18

You do indeed. Well done!

Bob

ps

I made it

which shows what I mean about many answers being possible.

Last edited by Bob (2011-01-03 05:05:47)

tstbs · 2023-08-08 11:58:39

If the vector equation of a line is an infinitely long vector, why cannot it be displaced like other vectors?
That is, does it represent a set of parallel lines?
If it is a vector it should only have magnitude and direction.

Bob · 2023-08-09 01:24:34

hello tstbs

Welcome to the forum.

Wow! That's an old thread you have discovered there. Takes me back a bit.

In 3D one way to have the equation of a line would be to go to a point on the line then a variable amount in the drection of the line eg.

If you change a you could get a parallel line.

The line would be infinitely long in both directions.

Math Is Fun Forum

#26 2011-01-01 03:08:46

Re: Vector Equations

#27 2011-01-01 04:49:15

Re: Vector Equations

#28 2011-01-01 04:51:25

Re: Vector Equations

#29 2011-01-01 05:47:24

Re: Vector Equations

#30 2011-01-01 06:17:15

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#31 2011-01-01 06:30:20

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#32 2011-01-01 22:41:23

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#33 2011-01-02 01:49:17

Re: Vector Equations

#34 2011-01-02 01:53:30

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#35 2011-01-02 01:55:25

Re: Vector Equations

#36 2011-01-02 02:03:43

Re: Vector Equations

#37 2011-01-02 06:05:36

Re: Vector Equations

#38 2011-01-02 07:35:10

Re: Vector Equations

#39 2011-01-02 08:52:15

Re: Vector Equations

#40 2011-01-03 02:06:35

Re: Vector Equations

#41 2011-01-03 02:34:48

Re: Vector Equations

#42 2011-01-03 04:02:24

Re: Vector Equations

#43 2011-01-03 04:14:16

Re: Vector Equations

#44 2011-01-03 04:48:18

Re: Vector Equations

#45 2023-08-08 11:58:39

Re: Vector Equations

#46 2023-08-09 01:24:34

Re: Vector Equations

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