eyal@fir.canberra.edu.au (Eyal Lebedinsky) writes:

>Hi,
> This should be simple but I can do without figuring it out myself.
>Basically I have two objects and I need to find thei closure rate. So if
>the vector between them is A and their relative velocity is V then I want
>to find that part of B along A. A unit vector for A should be sufficient
>and I want to do NO trig, just simple mul/add. I think this should be
>simple (maybe just an inner product of the unit vector and V?) but am
>not sure.

>Thanks in advance.

Yes, it is simple. So simple, I keep looking in my lin. alg. book for
it. (And I do not have that here...)

-------*---------------> Vector A
\      +
 \     +
  \    +
   \   +
    \  +
     \ +
      _| Vector B

It was something like: a * aTb / aTa (From base to *)
              [ x ]
(In which a = [ y ] , so aT = [ x y z ], thus aTb (and aTa) is a number, not
              [ z ]
 a vector or a matrix)
                                 [ 4 ]         [  1 ]
I'll run a check in 2d:  Say A = [ 0 ] and B = [ -3 ].

Then aTb = (4*1) + (0* -3 ) = 4
aTa = 16, so aTb/aTa = 1/4. 
                    [1]
So, a * aTb / aTa = [0], and that's correct.

geesh, I still remember! And that straight out of my head. No references.
Great.

BTW, If you want the vector perpendicular to A, you calculate of course:

b - ( a * aTb / aTa ).

Hope this helps,

Wouter Liefting
wlieftin@cs.vu.nl