Hi Bob!!!
Good to know you're still on these forums, I remember you helped me get past a lot of challenging problems last year.
I think so, is this the way to do this?  Sorry for the bad notation i'll have to go over the LaTeX post again...
x=p+t(q-p)
(x,y,z)2,4,-1)+t((5,0,7)-(2,4,-1))
(x,y,z)2,4,-1)+t(3,-4,8)
x=2+3t
y=4-4t
(x,y)2+3t,4-4t)
Thanks a lot

hi Deon588

Yes, you have done that well!    (This emoticon is meant to be there!)

will do nicely.

so you have

x = 2 + 3t
y = 4 - 4t

and
z = -1 + 8t

Don't ignore the z component ... rather make it equal to zero

0 = -1 +8t

solve for t

That fixes the value of t and so you can put back into the x and y equations to get the point as three numbers.

Bob

hi Deon588

That's a tough question.  I'd hate to give the 'wrong' answer.

Would you be able to post a question or two so I can see what sort of thing they're asking?

Bob

hi Deon588

Ok, that's a relief for me.  I had thought I might be given questions beyond my knowledge ... then I'd have been in trouble.  But these look OK although I cannot see my way to the end of Q3.  I'll have to write stuff down.

EDIT:  Actually I can see it comes from the same theorem as Q2. so Q3 is ok too.

But these are just basic complex theory, not eigenvectors.  So do you have basic complex on the exam or is it just during linear algebra, eigenvectors; in which case I'd still like to see a question on that.  On the basis of what I've seen and you have said about the time factor, it seems you could take the risk and leave out eigenvectors.  But, please don't blame me if lots come up.  Don't forget, you're the one looking at the past papers.

Bob

hi Deon588,

That's OK.  The questions you posted don't look too bad so if you want to do the theory I'm here to help if needed.

http://www.mathsisfun.com/numbers/complex-numbers.html

http://en.wikipedia.org/wiki/De_Moivre's_formula

Bob