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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Hi,

I'm not sure how to do a question which I have on vector equations - vectors have never been my strong point .

The points *A* and *B* have position vectors

And

Respectively. The points *A* and *B* are transformed by the linear transformation *T* to the points *A'* and *B'* respectively.

The transformation *T* is represented by the matrix **T**, where

a) Find the position vectors of *A'* and *B'*.

I was able to work out that:

.And

.b) Hence find a vector equation of the line *A'B'*.

Here is where I'm completely stuck . I haven't done any of the vector units which precede this one.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,475

Hi Au101;

Take a look here and see if this helps you.

http://www.revisesmart.co.uk/maths/core … -line.html

Here is an interesting vid that covers some basics of it.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

hi Au101,

Back so soon ?

Part (a) all correct!

Part (b)

A vector equation is similar to the cartesian equation y = mx + c

In y = mx + c you have an independant variable = x; a direction = m; a fixed point = c ( strictly 0,c ) and dependant variable = y

For vectors

lambda is the independant variable and is a scalar number; A'B' is the direction; OA' is the fixed point (you could use OB'); and r is the dependant variable.

So find A'B' and substitute in the general vector equation.

To get to any point on the line, first go from the origin to A'(OA') , and then take a variable trip along the vector A'B' (scalar.A'B')

To test if your equation works, find the lambda that makes r = OB, then find the lambda that finds the mid-point of A'B'

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Hmmm thanks Bobby, so could I then say:

Oh hi Bob Bundy, I didn't see your reply there, you replied just as I was, thanks a lot

*Last edited by Au101 (2010-12-29 08:02:50)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

hi Au101,

Right answer! Well done.

You could still do these bits:

To test if your equation works, find the lambda that makes r = OB, then find the lambda that finds the mid-point of A'B'

Bob

*Last edited by bob bundy (2010-12-29 08:36:38)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Certainly I shall, if only out of interest, I was especially interested in whether my layout was appropriate. Thanks, once again

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,475

HIi Au101;

I am looking at some of the videos over there they are pretty good. What did you think of them?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Hi Bobbym, I agree, instructive and well laid out, good quality as well - sometimes they can be difficult to read - thanks

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

hi Au101

Layout:

Your layout is fine.

Just as you can have y =mx + c OR mx = y - c OR c = y - mx .......

so you can have your vector equation in a variety of forms.

And you can go to a different point on the line

That's OB' followed by a 'different' lambda

Or even:

These are all the same line. Makes it hell to mark when every student may submit a correct, but different, equation.

For this reason you may also find your answer looks different from the book answer.

That's why it is worth trying to generate particular points on the line just to check it works.

If you get one answer and the book another, they are the same when they generate the same points.

Bob

*Last edited by bob bundy (2010-12-29 09:32:40)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,475

Sometimes it is a lot easier to learn by watching some videos rather than going through a book. They supplememnt each other well.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

Hi Au101 and bobbym

What follows is my first ever attempt at this.

This Shockwave version is 1.24M and the conversion has killed the resolution.

http://www.bundy.demon.co.uk/images/vectorline.swf

Try the avi version

http://www.bundy.demon.co.uk/images/vectorline.avi

It's now 2.8M and you'll need Windows Media Player.

Comments for improvements welcomed.

Bob

*Last edited by bob bundy (2010-12-30 01:33:26)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,475

Hi Bob;

Nice!

The program you used to make the movie? Some of them can monitor a portion of the screen, Snagit can I think. In other words they make a movie out of a selected area on the screen. What program are you using? Does it have that capability?

**In mathematics, you don't understand things. You just get used to them.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

hi bobbym,

It's a free download called Camstudio and I've just discovered how to do exactly that. New version is now uploaded and it is only 2.8M.

Bob

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

That's very nice . I think I get what the equation means now, thanks!!! . I am, however, stumped by a new, but similar form of question, I have a worked example, but as I've alluded to this is the last chapter of a unit which is way ahead of anything I've done and I don't really understand some of the ideas, perhaps someone could explain what to do and why?

The transformation

Is represented by the matrix **T**, where

The plane

is transformed by *T* to the plane

The plane

Has Cartesian equation

Find a Cartesian equation of

.*Last edited by Au101 (2010-12-30 02:29:41)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

Hi Au101,

I've got the bug for videos now. I've just put a sine curve video link into Teaching Resources. If you get a chance, please have a look. That may be it for now, as these vids have used up all my allocated web space!

Now to your problem. You seem to have moved a long way in 24 hours!

Planes have the form ax + by + cz = 0 so that they have two degrees of freedom. If you think about the line, it had just one because you could only choose the value of lambda. Now with a plane you have two choices, say, x and y; then z is determined.

The vector

is perpendicular to the plane. (Do you need that explained because it is another post in itself).

edit: I think what I said here was wrong. See post 20 for correction.

Bob

*Last edited by bob bundy (2010-12-30 06:16:20)*

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Hmmm such an explanation may indeed be helpful - the reason why I've moved so far in so short a time is that my textbook assumes knowledge of chapters I haven't yet covered. Also, I'm sorry to hear that you're out of bandwidth but I would be happy to have a look, I don't quite know where you mean by 'teaching resources' however?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

http://www.mathisfunforum.com/viewtopic.php?id=14858

B

explanation in prep Do you know what is meant by the dot product (also called the scalar product) ?

*Last edited by bob bundy (2010-12-30 02:54:30)*

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Ah, thank you. I haven't covered dot product, but I have looked at it before and I'm fairly sure that I know what it does and how to use it.

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Ooooh that is, indeed, very nice, I must say, it is very clear and well-done. It is good to see it explained this way - of course I've come across it many times - but it was originally explained to me arguably less well.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

hi Au101,

Thanks for the comments.

I've finished playing about with these posts now. Either they are right or my brain will explode!

There's so much I'm going to split this post in two. First a look at the scalar product; the general equation for a plane; and then the particular case where the plane goes through the origin.

But I now don't think I was right to assume that the perpendicular vectors get transformed the same way so I'll put a proper method into the next post.

Three stages:

(i) The dot product.

If a and b are vectors then a.b is defined as |a|.|b|.cos θ = a1b1 + a2b2 + a3c3.

(θ is the angle between the vectors)

In particular if the vectors are perpendicular then a.b = 0

(ii) The general equation of a plane.

see picture below

To get the position vector of D, find how much you need to go in the 'a' direction and the 'b' direction to get from C to D, then do OC + CD.

Now suppose that vector 'n' is perpendicular to the plane. That means it is prependicular to every vector in the plane.

Then

But a.n = b.n = 0 as these vectors are perpendicular.

So r.n = c.n = some constant as c and n are fixed.

So if x,y and z are the components of r and n1, n2 and n3 are the components of n

x.n1 + y.n2 + z.n3 = constant.

This can be used to define the plane eg 3x + 4y - 6z = 27

(iii) Planes that go through (0,0)

Matrix multiplication can only be used for transformations that leave the origin invariant

ie. M . 0 = 0

So your transform must be one of these otherwise they couldn't set this question. And we can see that the origin is a point in the plane as x = y = z = 0 fits.

So the plane constant must be zero

So

is a possible vector for n, perpendicular to the plane. (Or any multiple of n)

Bob

*Last edited by bob bundy (2010-12-30 07:20:17)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

hi

post part two

steps: (i) find two vectors in the first plane (ii) transform them (iii) form an equation for the plane that contains these.

step (i)

By inspection these vectors will both lie in the plane

step (ii)

Transform these to get

step (iii)

so the equation is:

so

A x 6 =>

Then add C

also add B and C

E x 8 =>

add D

Bob

*Last edited by bob bundy (2010-12-30 07:15:11)*

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Ooooh thanks very much for all of your hard work Bob, I think I've genuinely got it now, the only thing I don't think I quite follow, ironically, is the first step which, I assume, is the easier step, where did you get those vectors from?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,463

hi Au101,

The equation was

So I wanted two vectors (chosing x, y and z) that would fit that equation.

I just spotted x=y=z=1 as one possibility and then x = 1 y = 0 and z = -1 as another.

I was pleased to choose those because the numbers were easy to work with.

But any three numbers that fit the equation would do for a vector in the plane:

Let's have x = 39 and y = 22. Then z = 44 - 39 = 5.

That's what I meant before by two degrees of freedom.

Choose any two and that fixes the third.

So z = -3, y = 100 => x = 200 + 3 = 203. And so on.

But 1,1,1 and 1,0,-1 were easier to work with.

The only choice you cannot make is to have two vectors that are parallel, eg. x = y = z = 1 for one choice and x = y = z = 2 for the other. They are valid choices but you cannot solve the problem because you don't get enough independent equations to solve for mu and lambda.

Look back to my diagram for the plane. If vectors a and b are parallel you would not be able to get to all possible places for D.

The vectors have to be 'independent' (meaning one cannot be made by multiplying the other by a fixed amount). I expect you'll meet that idea again in this module.

Bob

*Last edited by bob bundy (2010-12-30 08:54:27)*

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

Ahhhhhhh I see, thanks!!!

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**Au101****Member**- Registered: 2010-12-01
- Posts: 286

*Last edited by Au101 (2010-12-31 04:13:20)*

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