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QUESTION 2 b - last years paper

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Anonymous User
QUESTION 2 b - last years paper
October 02, 2006 10:31AM
Can anyone solve Q2b on last years paper. If so please help me. I got no idea what to do here. I tried reading up on planes in chapter 4 - but good luck because it still doesnt fix my problem.

Please help guys1
Anonymous User
Re: QUESTION 2 b - last years paper
October 02, 2006 05:27PM
I really dont think they are going to ask this question in the exam
as it was not in the assignments ,exta tutorial letter or in appendix a or b.

But I will email the lecturers

Anonymous User
Re: QUESTION 2 b - last years paper
October 04, 2006 11:49AM
I know, probably not. But in the odd case they do, i just want to know how to solve it.
Anonymous User
Re: QUESTION 2 b - last years paper
October 04, 2006 12:05PM
Could you solve Q2(a)?
Anonymous User
Re: QUESTION 2 b - last years paper
October 04, 2006 06:54PM
HAve no idea how to solve it.
Re: QUESTION 2 b - last years paper
October 05, 2006 08:03AM
I think I got it.

show that the first three vectors are linearly independent. this means that they are not a linear combination of each other, which means that they do not all lie on the same line. which means that they form a plane. then show that the last point is a linear combination of the previous 3.

I believe linear combinations and planes are part of the subject matter so I guess it is possible for it to be asked in the exam.
Anonymous User
Re: QUESTION 2 b - last years paper
October 06, 2006 08:14AM
Hey I also thought that would be the way to do it. Thanks for your help! It makes sense to do it that way. Im going to go test it out. Thanks again!
Anonymous User
Re: QUESTION 2 b - last years paper
October 06, 2006 04:31PM
Will the maths in the past exam paper (for example question 2b) be in the exam?

Yes, there will be questions on maths in the exam. They will be similar to the questions in last years paper, but not the same.
KH
Re: QUESTION 2 b - last years paper
October 15, 2006 08:11PM
Can we not use the homogeneous points and then calculate the determinant?
Re: QUESTION 2 b - last years paper
October 16, 2006 03:37PM
maybe. please explain how you would do that and why it would work though? the determinant seems like magic to me - it gets used for all kinds of strange things.
Re: QUESTION 2 b - last years paper
October 16, 2006 04:58PM
The 4 points are (0,1,2),(1,2,0),(2,0,1) and (0,4,0)
For homogeneous points, the fourth value is 1 as opposed to 0 for vectors.
You then have a 4x4 matrix

0 1 2 0
1 2 0 4
2 0 1 0
1 1 1 1

If you multiply it out you get 0+0+16+0-0-8-0-0=8
This means the points are linearly dependent and threfore are planar
avatar Re: QUESTION 2 b - last years paper
October 19, 2006 12:30AM
Chaospixel wrote:
>show that the first three vectors are linearly independent. this
>means that they are not a linear combination of each other, which
>means that they do not all lie on the same line. which means that
>they form a plane. then show that the last point is a linear
>combination of the previous 3.

All that you get when you prove that 3 vectors are linearly
independent is that they cannot be expressed in terms
of each other i.e. they are orthogonal to each other ...
like the standard basis in R3 ... the X, Y and Z axis
are all linearly independent.

Showing that the last vector is a linear combination
of any of the others just shows that it can be expressed
in terms of the others; a normal to the plane can be
expressed the same way but does not necessarily form
part of the plane.

Sherlym wrote:
>Can we not use the homogeneous points and then calculate the
>determinant?
...
>The 4 points are (0,1,2),(1,2,0),(2,0,1) and (0,4,0)
>For homogeneous points, the fourth value is 1 as opposed to 0
>for vectors.
>You then have a 4x4 matrix
>
>0 1 2 0
>1 2 0 4
>2 0 1 0
>1 1 1 1

This is wrong, you know. You can only express 3 coordinates
in a 4x4 homogenous coordinate system.

>
>If you multiply it out you get 0+0+16+0-0-8-0-0=8
>This means the points are linearly dependent and threfore are
>planar

I'm having trouble following your logic here. Can you
perhaps explain further? What did you multiply the
matrix with (the transpose? the identity?) and why?

The reason I ask is because I've not found an easy
solution to this (although lots of tedious and long
solutions have come my way smile.

btw: to prove a set of 3 vectors are linearly independent,
you merely have to solve the set of simultaneous equations
(like you learned in standard 7) consisting of the three
equations of the vectors in x, y and z unknowns.

Getting 0 for all 3 means that none of them can be expressed
in terms of the others, getting a single non-zero result
(impossible to get only *one* non-zero result - you would
get at least two) means that one of the vectors can be expressed
in terms of one or more of the other vectors.

Good luck

--
Learn something new - updated weekly
avatar Re: QUESTION 2 b - last years paper
October 19, 2006 12:58AM
(Editing this, it's little terse without
explanation, so adding some explanation in).

Try this:
You've got 4 points, A B C and D.

Subtract B from A to give vector U
Subtract C from A to give vector V
Subtract D from A to give vector W

Find the cross-product of UxV; call the result vector R

Find the dot-product of R and W, if the scalar
result is 0, then all 4 points lie on the same plane,
else they are not all on the same plane.

Basically, what you are doing is constructing
3 vectors out of the four points, then constructing
a 4th vector that is orthogonal to two of your vectors
(which means it must be at 90degrees to both of them).

Then you finally work out the angle between your
new vector and the last of the original 3 you created.
If the angle between your 90degree vector and the
last vector is 90 (dot-product==0 means that the
angle is 90degrees), then you've answered the question.

Pretty nifty, eh smile Easy to remember without needing
big words like "determinants" (which still give me
nightmaressmile

Anyway, off to sleep now ...

Good luck all
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