Vector help
 

I need help trying to understand a problem in a maths question involving vectors (which I do have the answer to). I know there a quite a few mathmaticians around here so I plead for some help.

I know that a.b means the dot product of those two vectors but the working makes no sence to me.
The problem is as follows:
The answer I am given is:
Those cos0 are supposed to be the beta symbol thing, a 0 with a cross in the middle)
Now what I want to know is where did what does c.d have to do with anything? What are t and s and how do they know where the lines intersect?

Sorry about this but I'm confused and a maths dunce.

Thats the kinda stuff I'm doin for matsh atm but I havent done that yet, sorry...
What level maths is it? University, school etc? :)

Hmm.I dont remember much but here goes.

s & t are scalars.

To find the vector eqaution you have to use the standard form.
To find the angle we use c and d because of the scalars.The point of intersection is found by solving the equation.We get the scalars as -2 and 1.

all I can say.. gulp.. goodluck phily baby!

Thanks dam, that's kinda set me on the right track. This is university maths for my graphics course. Which is good because the last time i did maths was at GCSE where I struggeld. Things are starting to click now however :).

Right..

the c.d is used to calculate the angle because when two lines intercept, you can use the formula

cos (theta) = (c.d)/((magnitude of c) x (magnitude of d))

code:


cos theta =   c.d 

             |c||d|

therefore...

c.d = (0x2)+(1x1)+(-1x-1) = 2
mag c = sqrt(0²+1²+(-1)²)) = sqrt(2)
mag d = sqrt(2²+1²+(-1)²)) = sqrt(6)

therefore cos(theta) = 2/(sqrt(2)*sqrt(6) = .57735

therefore theta = arccos(.57735) = 54.74°

more to come if I can remember it (it's been a while)

oh and the reason you use c and d instead of a or b is becuase if you consider it to be like a graph, the a and b just represent points whereas c and d represent the gradient (this is a crap explaination not sure if it will help).

take the line r=a+tc for example..

effectively you start at point a, you then go off at angle c for a distance that is a multiple t of the magnitude of c. from this you can calculate your position r


for calculating the intercept point...

r=a+tc
r=b+sd

therefore

a + tc = b + sd

putting in numbers...

code:


[ 1] [ 0]  [ 5] [ 2]

[ 2]+[ 1]t=[ 5]+[ 1]d

[-1] [-1]  [-4] [-1]

from the top one

1+0t = 5+2s
1 = 5+2s
-4 = 2s
therefore s=-2

then using any of the other lines ... (I have used the second

2+t = 5+d
2+t = 5+(-2)
2+t = 3
therefore t = 1


putting s and into the 1st equation r=bs+d
you get r=[1,3,-2]

Right I'm off to buy Mathematical Methods for Science Students for some extra help. Well I'll go see if it's better than the book I've got at the moment.

Thanks pAk, I've got upto figuring the angle, I take a long time to grasp some maths. :)

added a load more above, just in case you don't notice

I don't understand the point of the last question, but

n=b-a = [5,5,-4]-[1.,2,-1] = [4,3,-3]

then using the dot product again

r.n=k

(1x4)+(3x3)+(-2x-3)
=4+9+6
=19

God, I have a headache now. HAvn't done any of that in years.

Must get back to revision...

anyone know anything about laplace transforms?

WOW! Thanks man that's helped a lot. I got me new book too so am now feeling a little thin in the wallet.

Its worth the cash.Better than trying to figure it all out on your own.

Maths is one of those things I've always wanted to be good at but never have been. Now I think it's because I haven't put the effort in for a long while and make too many silly mistakes like missing off +/- signs. I'm now going to try get my act together before the end of the year.

I just know my graphics coursework (that made no secnce at all to me) will make perfect sence by the end of the summer, by which time it's too late.

So far I can find the angle and the vector P but the last bit I don't get, time to e-mail the lecturer...