I need some math help

[The Mullinator]

No im not asking you guys to explain something to me or help with my homework so don't worry.

Right now im in a grade 12 math course called "Geometry and discrete math" and I need to boost my mark. I can understand what is being taught but only barely and I feel the best way I can understand it better is to find a decent computer program that can help teach what is in the course to me, or find a really good website to help me out.

Right now we are working with vectors, adding them, finding the dot product between 2 vectors and finding the cross product between vectors, etc.

Here is a basic description of the course if your interested:

The Geometry and Discrete Mathematics (MGA4U) course enables students to broaden mathematical knowledge and skills related to abstract mathematical topics and to the solving of complex problems. Students will solve problems involving geometric and Cartesian vectors, and intersections of lines and planes in three-space. They will also develop an understanding of proof, using deductive, algebraic, vector, and indirect methods. Students will solve problems involving counting techniques and prove results using mathematical induction.

If anyone knows of a decent program or website to help me out it would be extremely usefull to me (I hope).

Thanks.

 

[manny_c44]

Don't you have a book?? Thats usually helpful....

Generally math websites are not helpful at all since they are about college math and abstract concepts.

 

[The Mullinator]

I have the text book but it is absolute crap, you can't learn anything from it.

Alot of other books have looked hopeful but I find I learn better working from and reading from a computer than a book. Don't ask why.

 

[LoneDeranger]

Yeah math books and math websites (in general) suck. Your best bet is to get together with some people from your class and try to figure it out.

 

[manny_c44]

Well for vectors you just break them down into components and add them as x and y (and z if working in 3d planes) and then put back into vector. Or is this something different?

 

[The Mullinator]

Originally posted by manny_c44
Well for vectors you just break them down into components and add them as x and y (and z if working in 3d planes) and then put back into vector. Or is this something different?

Thats exactly what it is, but see what we are expected to do is a bit more complicated than that.

Here is an example question from the text book:

"Use the cross prduct to find the area of the triangle whose vertices are at coordinates A(-7,3,5) B(3,1,2) and C(2,-6,6)"

Or we have to prove an equation is true:
"Prove [a x (b+c) = a x b + a x c]
a,b,c are vectors"

I can answer both these questions but not without taking a long time (too long for tests), and practice from the text book doesn't seem to help me much. I think its just a lack of a thorough understanding of the concepts at work.

If my only real hope is with getting together with classmates then thats what I'll have to do. Although I would still prefer a way of learning on my own.

 

[nw909]

Ah yes, grades 10-12 where they teach you math that you will never use in real life but still have the guts to fail you.

 

[Hallucinogen]

I hate math with passion.

 

[The Mullinator]

Originally posted by nw909
Ah yes, grades 10-12 where they teach you math that you will never use in real life but still have the guts to fail you.

Unfortunatly I need it to get into computer science for university. ;(

Then as it turns out they don't need the Computer Science courses, the ones I get like 99% in (im not joking). They really know how to screw me up, "for the Computer Science course you need the hardest highschool course in the province where you get low marks, but you don't need the highschool Computer Science course where you get 99%. :flame: :flame: :flame:

 

[nw909]

University isn't real life!

 

[manny_c44]

Explain to me what this "cross-product" is.

 

[The Mullinator]

Originally posted by manny_c44
Explain to me what this "cross-product" is.

Here is a general definition:
"A cross product is a vector perpendicular to the plane containing a pair of non-collinear vectors, such as (3,4,-1) and (2,5,3)."

Its difficult to explain without using diagrams.

 

[Brian Damage]

Okay, we're getting somewhere. Now define "Non-Co-linear Vectors".

EDIT: Perhaps it means non-parallel.

Math doesn't suck, math RULES, but the way that they teach it bites.

 

[The Mullinator]

"Non-Co-linear vectors": These are vectors that are NOT parrallel to each other.

These two vectors ARE Co-linear: (1,1,1) (2,2,2)

EDIT: Yes your right.

 

[Brian Damage]

How can you have a plane between two non-parallell vectors? Something fishy's going on.

EDIT: I think we need a diagram. Got a scanner?

 

[The Mullinator]

Originally posted by Brian Damage
How can you have a plane between two non-parallell vectors? Something fishy's going on.

EDIT: I think we need a diagram. Got a scanner?

No I don't have a scanner sorry.

You see the plane the 2 vectors creates is not the x,y, or z plane. Say you have 2 vectors in 3d-space, if you were to draw a line between the end point of each vector you would get a line. That line is also the plane that is created.

did that help?

 

[manny_c44]

How would a single line be considered a plane? They would be points lying on a seperate plane.

 

[Brian Damage]

Plane = 3D

Line = 2D

We really need a diagram. Got a digital camera?

 

[manny_c44]

Actually
line =1d
Plane=2d

Line can be on a plane but not the other way around. Its difficult to teach math like this without visual aid or a piece of paper to draw on.

I suggest you re-read the book's explanation very carefully multiple times. Or *gasp* ask your teacher.

 

[Dux]

If:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Is represented as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
22 23 24 25 26
Then:
H-A-R-D-W-O-R-K
8+1+18+4+23+15+18+11 = 98%
and
K-N-O-W-L-E-D-G-E
11+14+15+23+12+5+4+7+5 = 96%

No, I've no idea why I posted this either.

 

[Brian Damage]

Originally posted by manny_c44
Actually
line =1d
Plane=2d

Line can be on a plane but not the other way around. Its difficult to teach math like this without visual aid or a piece of paper to draw on.

I suggest you re-read the book's explanation very carefully multiple times. Or *gasp* ask your teacher.

Yeah, I know lines are 1D and planes are 2D, but we're talking about lines and planes in 3D space.

But anyway, even if you have two lines, there's no gaurantee you can get a plane from them. They'd need to be parallel in some fashion, wouldn't they? Otherwise you'd end up with a curved surface, for which you'd need to define a specific 3D point to get a perpendicularity from.

 

[The Mullinator]

Sorry its difficult to explain.

here is a picture I whipped up to try and explain

The black lines are the xyz axis, the red are to vectors, and the blue represents the plane created by the two vectors.

Thats the best I can do right now since its late and I really have to sleep right now. See you guys tommorow.

 

[dfc05]

Originally posted by The Mullinator

Here is an example question from the text book:

"Use the cross prduct to find the area of the triangle whose vertices are at coordinates A(-7,3,5) B(3,1,2) and C(2,-6,6)"

*puts down history book and digs up pre-cal notes*
umm, yeah this is one of my test problems...do i understand what i was doing...no, not really but maybe it will help...
"given A(1,2,3) B(1,5,7) and C(3,3,1) -- a)find cosA and sinA; b)find the area of triangleABC"
so i found vector AB=(0,3,4) and vector AC = (2,1,-2)
[letting * mean dot product) and u=vectAB and v=vectAC and |u|=magnitude (length) of the vector
then cos A=(u*v)/(|u| |v|)
cos A=[(0,3,4)*(2,1,-2)]/[sqrt(0+9+16)xsqrt(4+1+4)]
cos A=(0+3-8)/(5x3)
cos A=-1/3
(sin A)^2 + (cos A)^2=1 so plug in cos A...
Sin A= (2xsqrt 2)/3
ok now finding area:
Area = (1/2)bc(sinA) where b=length of AB, c=length of AC
so plug stuff in and get 5xsqrt 2

ok so i didn't use cross product...umm, are you going to have to use a particular method on your test?

 

[dfc05]

ok i found one using cross product
points are P(0,2,3) Q(1,1,1) R(2,1,3)
so find vectors:
vect PQ = (1,-1,-2) and vect PR = (2,-1,0)
so cross prdct for PQ and PR:
|i j k| i j
|1 -1 -2|1 -1
|2 -1 0|2 -1
=-4j-k-(-2k+2i)
=-2i-4j+k
=(-2,-4,1)
then to find area of triangle
Area=(1/2)|(vectPQ)X(vectPR)| <--half the magnitude of the cross prdt
=sqrt(4+16+1) /2
=(sqrt 21)/2

and i still don't really understand what i was doing in that problem.... ugh

edit: spacing is messed up in finding crss prdt; the columns should be aligned

 

[jat]

I remember that crap, I was pretty far in my class till it got to A level where i wasn't putting in the effort. Basically this is what i did, there are a certian type of questions they can ask you in each math section. All you got to do is make a unviersal method to solving one of the types of vector questions u get, learn that and ur sorted. This way u dont need to understand the rules...
Do this for each type and learn how to apply it correctly, if this doesnt work, forget sites, go to a teacher or ur mates and just attempt em or ur not gonna get anywhere...
I cba reading all posts so dont flame me, just trying to help.....

 

[The Mullinator]

Originally posted by jat
I remember that crap, I was pretty far in my class till it got to A level where i wasn't putting in the effort. Basically this is what i did, there are a certian type of questions they can ask you in each math section. All you got to do is make a unviersal method to solving one of the types of vector questions u get, learn that and ur sorted. This way u dont need to understand the rules...
Do this for each type and learn how to apply it correctly, if this doesnt work, forget sites, go to a teacher or ur mates and just attempt em or ur not gonna get anywhere...
I cba reading all posts so dont flame me, just trying to help.....

Thats the way I have been doing it up until now, and thats why my marks are low. See in this course they try and teach us to figure something new out on our own, we HAVE to know the rules to figure out how to answer a question that involves things we havn't really learned yet. I would be getting in the high 80's to low 90's if it was simply learning how to answer a question and then just use that method on the test to answer the same kind of question. Instead we have to do that as well as using the rules to make up our own functions and equations and methods to answer questions we were not taught how to do.

dfc05, that is the method I use, although I also don't really understand how it all works out. :dozey:

 

[Echelon]

Originally posted by nw909
Ah yes, grades 10-12 where they teach you math that you will never use in real life but still have the guts to fail you.

vectors are pretty important u need them for many technical professions, but don'T ask me for what exactly, i also am in the 12th grade

 

[simmo]

I must be the the most useless person at maths on these forums...I cant understand a word you lot are saying lol :cheese:

 

[dfc05]

the best guess i can come up with as to how it works, is that maybe the magnitude of the cross product gives the area of the parallelogram that would be formed if you added the two vectors and used them as two sides of the parallelogram, and then to find the area of the triangle you'd take half the area of the parallelogram. that's just a guess though; i'm not even sure what cross product means (i don't think i ever learned an actual definition that made any sense to me) and i'd probably have to see some sort of diagram to figure out what it's supposed to be doing.

 

[The Mullinator]

Originally posted by dfc05
the best guess i can come up with as to how it works, is that maybe the magnitude of the cross product gives the area of the parallelogram that would be formed if you added the two vectors and used them as two sides of the parallelogram, and then to find the area of the triangle you'd take half the area of the parallelogram. that's just a guess though; i'm not even sure what cross product means (i don't think i ever learned an actual definition that made any sense to me) and i'd probably have to see some sort of diagram to figure out what it's supposed to be doing.

I think your right, I remember my teacher mentioning something about parrallelograms in order to find the area of a triangle not too long ago.

As for a clear definition of a cross product i'm not sure of what it may be myself.

 

[gantrithor_5]

My head hurts.

 

[dfc05]

*puts down English essay assignment, and looks in calculus book*

ok the book says that:
1) ||u X v|| = area of parallelogram having u and v as adjacent sides (|| means magnitude)
and it proves that by saying
2) ||u X v||=||u|| ||v|| sin A (where A is the angle b/w u & v) and then showing a diagram with the parallelogram where the height is ||v|| sin A.
they also proved the second statement by replacing sin with its equivalent in cos and doing weird stuff with the cosine thing using dot products..... (of course they would also have to prove the whole dot product property too... ugh)
and they also have a diagram showing what the cross product is (diagram didn't really help, but i'll stick it on)

i hope it helps a little; i still don't get it though, lol