Math Is Fun Forum

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

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

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

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

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

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

hi Au101,

I'm very impressed.  I'd want to take an exam too.

I'd better look up linear transforms.

Bob

hi loida and bobbym,

I'm getting 136/3 times pi.  ??

loida:  Where did your equation come from?  Why 2 pi ?

Bob

Hi MIF,

Thank you for that comment.  I still have the brain cells working on this.  My Sketchpad files are up to version 7 now.  I'm hoping there's a 'straight pythag. with 1 variable' solution lurking in there somewhere. 

Happy New Year!

Bob

hi loida

edit complete

I always start these with a diagram:

The line y = 1 forms the lower constraint;  x = 5 is the right hand constraint; and y = x forms the top constraint.

Note:  It is rotated around the y axis.

The dotted rectangle represents a 'typical' thin strip with width (5-y) and thickness dy.

So that 'rectangle', (5 - y) by dy, is rotated to make a disc (with hole) with volume pi (5 squared - y squared) times thickness

Now sum these up across the y direction:

Sorry, this does not give your answer.

Bob

hi Au101

Yes, they're really nasty!  What did the problem start with?

edit :  Actually, I think I've spotted a short cut.  Is c = 3 by any chance?

Bob

hi Au101

I've never actually called it 'set builder notation, but I know what you mean.

Have a look at:

http://en.wikipedia.org/wiki/Set-builder_notation

For me, it's just a sentence but written in the language of maths.

So if you want to say 'S is the set of matrices where the determinant is zero,  for all values of a, b and c in the real numbers'

how about

when

               (You will have to substitute the correct matrix here.)

Then

The format is

<name of set> = { <restrictions on variables> such that <formula defining set> }

Bob