Does anybody know where I can get past year papers for this subject?

Thanks]]>

-------------------

I studied to realise that knowledge humbles one.....]]>

Yes, you can. Don't be surprised if you get abruptly blown off though.

The lecturers, since first year, have always stressed that they don't

have anything to do with admin; there is no reason to believe that

the rules have magically changed for this module without any indicator

as such form the lecturers themselves.

OTOH, do you *really* think that it is a good idea to annoy the

lecturers while they are marking our papers? If you do, then please

give them your student number so they won't take out the annoyance

you've caused on the paper that they are marking at that moment in

time. Let them be annoyed *with you*, not *with me*!!!]]>

just wanna know if the lectures entertain this by phone/email.

i'm so anxious abt this exam !!! and advanced systems development !!!]]>

Has anyone else received the above tutorial letter?

I got this today and it basically says that all students

in the college of human sciences will be allowed to

write a supp if they fail due to administrative errors

earlier in the year with the assignments.

Confirmation anyone? Does this apply to students doing

a Bsc. (computer science and information systems)?

I ask because I've not heard of the college of human sciences

before.

thanks]]>

On this note, good luck to those that are not finished with their exams yet ( I stll have 2 left)

Cheerio

Dariusz]]>

Can Ods please expand what he meant by "M"?)

As far as the exam itself, anyone else found the question

about the coloured cube odd? Which cube were they referring

to?

Did anyone here read my posting to a recent thread here

this morning at around 1:00AM? The answer to the question

of finding the co-planarity of 4 points was basically

given in my posting.

I thought the exam was basically easy, considering that

the material was so complex, they could have really

nailed us with difficult questions (like computing

a phong reflection instead of just asking us to explain

what it is).

Doesn't mean I passed though - I was as unprepared as

one could possibly be ...

later]]>

How did youi find?? I failed certainly...not even a supp...]]>

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]]>

>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]]>

b = [1 1 0 1]T; f = [0 2 0 1]T

(a)

|1/2 0 0 0||1| |1/2|

| 0 1 0 0||1|=| 1 |

| 0 0 1 0||0| | 0 |

| 0 0 0 1||1| | 1 | and

|1/2 0 0 0||0| |0|

| 0 1 0 0||2|=|2|

| 0 0 1 0||0| |0|

| 0 0 0 1||1| |1|

(b)

sin(45) = 1/sqrt(2) but positive rotation is anti-clockwise and we're asked for

clockwise rotation: sin(360-45) = sin(360)cos(45)-cos(360)sin(45) = -sin(45)

= -1/sqrt(2)

cos(360-45) = sin(360)sin(45)+cos(360)cos(45) = +1/sqrt(2)

(so the hint is actually a trick - if you'd just remembered the signs wrong you

would have the calculated the answer correctly).

|1/sqrt(2) 0 -1/sqrt(2) 0||1/2| |1/(2sqrt(2))|

| 0 1 0 0|| 1 |=| 1 |

|1/sqrt(2) 0 1/sqrt(2) 0|| 0 | |1/(2sqrt(2))|

| 0 0 0 1|| 1 | | 1 | and

|1/sqrt(2) 0 -1/sqrt(2) 0||0| |0|

| 0 1 0 0||2|=|2|

|1/sqrt(2) 0 1/sqrt(2) 0||0| |0|

| 0 0 0 1||1| |1|

(c)

|1 0 0 0||1/(2sqrt(2))| | 1/(2sqrt(2)) |

|0 1 0 0|| 1 |=| 1 |

|0 0 1 2||1/(2sqrt(2))| |1/(2sqrt(2)) +2|

|0 0 0 1|| 1 | | 1 | and

|1 0 0 0||0| |0|

|0 1 0 0||2|=|2|

|0 0 1 2||0| |2|

|0 0 0 1||1| |1|]]>

What number do you think you have to use?]]>

a=(0,1,0); b=(0,1,1)

T=

1 0 0 x

0 0 1 z

0 0 0 1

Thus, T=

1 0 0 3.0

0 1 0 y

0 0 1 z

0 0 0 1

Now, I know you do a MULTIPLICATION btwn T and each given Point. My question is,

since the values of y and z are not given, do you regard them as 0 or 1?

Maybe i'm just plain stupid or somethin but i really need some clearity on this one. Anyone?]]>

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]]>

I have e-mailed them and asked them to post it.

Hopefully they will do this as soon as possible.

Regards

Arno]]>

send an email to cos340@osprey.unisa.ac.za]]>

I found most of the answers to the questions in the prescribed book on the internet.

Anyone interested in them can just give me their e-mail addresses and i will mail it to them.

Regards

Arno]]>

void earth()

{

float x, y, z, thet, phi;

float c = 3.14159/180.0;

for (phi = -90.0; phi <= 90.0; phi += 20.0)

{

glBegin(GL_QUAD_STRIP);

for (thet = -180.0; thet <= 180.0; thet += 20.0)

{

x = sin(c*thet)*cos(c*phi);

y = cos(c*thet)*cos(c*phi);

z = sin(c*phi);

glNormal3f(x, y, z); // just need 1 normal for flat shading

glVertex3f(x, y, z);

x = sin(c*thet) * cos(c*(phi+20.0));

y = cos(c*thet) * cos(c*(phi+20.0));

z = sin(c*(phi+20.0));

glVertex3f(x, y, z);

}

glEnd();

}

}

void display()

{

glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);

glLoadIdentity();

glRotatef(105, 1.0, 0.0, 0.0);

earth();

glFlush();

}

void reshape(int w, int h)

{

glViewport(0, 0, w, h);

glMatrixMode(GL_PROJECTION);

glLoadIdentity();

if (w <= h)

glOrtho(-2.0, 2.0, -2.0*(GLfloat)h/w, 2.0*(GLfloat)h/w, -10.0, 10.0);

else

glOrtho(-2.0*(GLfloat)w/h, 2.0*(GLfloat)w/h, -2.0, 2.0, -10.0, 10.0);

glMatrixMode(GL_MODELVIEW);

}

void init()

{

glEnable(GL_DEPTH_TEST);

// enable lighting

glEnable(GL_LIGHTING);

glEnable(GL_LIGHT0);

// set the shading model to flat

glShadeModel(GL_FLAT);

// position the light before any transformations take place

GLfloat pos[] = {0.0, 0.0, 2.0, 0.0};

glLightfv(GL_LIGHT0, GL_POSITION, pos);

glClearColor(1.0, 1.0, 1.0, 1.0);

// glColor3f(0.0, 0.0, 0.0); // optional

// glPolygonMode(GL_FRONT_AND_BACK, GL_FILL); // optional

glPolygonMode(GL_FRONT, GL_FILL);

}

int main(int argc, char* argv[])

{

glutInit(&argc, argv);

glutInitDisplayMode(GLUT_RGB | GLUT_DEPTH);

glutInitWindowSize(500, 500);

glutCreateWindow("sphere");

glutReshapeFunc(reshape);

glutDisplayFunc(display);

init();

glutMainLoop();

return 0;

}

]]>

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]]>