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Anonymous User
QUESTION 2 b - last years paper October 02, 2006 10:31AM |
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Anonymous User
Re: QUESTION 2 b - last years paper October 02, 2006 05:27PM |
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Anonymous User
Re: QUESTION 2 b - last years paper October 04, 2006 11:49AM |
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Anonymous User
Re: QUESTION 2 b - last years paper October 04, 2006 12:05PM |
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Anonymous User
Re: QUESTION 2 b - last years paper October 04, 2006 06:54PM |
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Re: QUESTION 2 b - last years paper October 05, 2006 08:03AM |
IP/Host: ---.enetonline.co.za Registered: 14 years ago Posts: 78 Rating: 0 |

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.

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 |
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Anonymous User
Re: QUESTION 2 b - last years paper October 06, 2006 04:31PM |
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Re: QUESTION 2 b - last years paper October 15, 2006 08:11PM |
IP/Host: ---.rdg.dial.mweb.co.za Registered: 12 years ago Posts: 5 Rating: 0 |

Re: QUESTION 2 b - last years paper October 16, 2006 03:37PM |
IP/Host: ---.enetonline.co.za Registered: 14 years ago Posts: 78 Rating: 0 |

Re: QUESTION 2 b - last years paper October 16, 2006 04:58PM |
IP/Host: ---.rdg.dial.mweb.co.za Registered: 12 years ago Posts: 5 Rating: 0 |

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

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

Re: QUESTION 2 b - last years paper October 19, 2006 12:30AM |
IP/Host: ---.saix.net Registered: 14 years ago Posts: 3,793 Rating: 0 |

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 .

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

>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 .

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

Re: QUESTION 2 b - last years paper October 19, 2006 12:58AM |
IP/Host: ---.saix.net Registered: 14 years ago Posts: 3,793 Rating: 0 |

(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 Easy to remember without needing

big words like "determinants" (which still give me

nightmares

Anyway, off to sleep now ...

Good luck all

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 Easy to remember without needing

big words like "determinants" (which still give me

nightmares

Anyway, off to sleep now ...

Good luck all

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