Maths help - vectors

|Show| · 26 Feb 2006 at 16:14

Just a quick question, and i thought this would be the best place for it.

Right, i'm trying to obtain the outward unit normal vector (n) of CHGF
using the formula -

Code:

[u]CH x GF[/u]
|CHxGF|

CH = (-2.5, 0, -2) and GF = (0, -10, 0)

i had a problem with CH x CF as surely thats 0 and then you can't calculate the normal if the first stage equals 0...

any ideas?

Minto · 26 Feb 2006 at 16:16

chxGf doesnt equal zero does it? i havent fully done it but the i component is -20i isnt it?

|Show| · 26 Feb 2006 at 16:18

well from what i did i got 0...but i could well be wrong...i've done too much on matrcies since vectors, so i'm not sure if all my methods are correct.

if its -20, then that'd make |CH x GF| - 20? (i've bracketed things in order to get possitive answers)

which then would make it
-20
20

=-1 :confused:

Minto · 26 Feb 2006 at 16:34

CH = (-2.5, 0, -2) and GF = (0, -10, 0)

Code:

CHxGF =   ¦ i     j   k ¦
          ¦-2.5  0  -2  ¦
          ¦  0  -10  0  ¦

= 25k -20i

|Show| · 26 Feb 2006 at 16:59

ah, now i see the problem... CF is 0,-10,0 (sorry) :eek:

GF is 2.5, 0, 2

and this is my problem -

Code:

CHxGF =   | i      j    k  |
           |-2.5   0    -2 |
           | 2.5   0     2  |

and don't they just cancel each other out and equal zero :confused:

and to add, the vectors for C, F and G are -
C = (2.5, 10, 8)
F = (2.5, 0, 8)
G = (0, 0, 6)

benjo · 26 Feb 2006 at 17:06

wait for it...

:\

There we go...

Naffa · 26 Feb 2006 at 17:50

benjo said:
wait for it...

:\

There we go...

ha.. ha..

Pickers · 26 Feb 2006 at 18:12

|Show| said:

These vectors are parallel - yes?

Arcade Fire · 26 Feb 2006 at 18:22

The vectors are parallel (CH-GF), so CH and GF lie on the same line, so the parallelogram CHGF is in fact a line, and thus has no unique outward-pointing unit normal.

The way that the maths tells you this is by giving you a zero result

Pickers · 26 Feb 2006 at 18:33

Arcade Fire said:
The vectors are parallel (CH-GF), so CH and GF lie on the same line, so the parallelogram CHGF is in fact a line, and thus has no unique outward-pointing unit normal.

The way that the maths tells you this is by giving you a zero result

OcUK - 1
Homework - 0.

|Show| · 26 Feb 2006 at 18:38

Arcade Fire said:
The vectors are parallel (CH-GF), so CH and GF lie on the same line, so the parallelogram CHGF is in fact a line, and thus has no unique outward-pointing unit normal.

The way that the maths tells you this is by giving you a zero result

ah, right...well this is the first part i guess in calculating the co-ordinates of the line which is perpendicular to the parallelogram CHGF. and it has a length of 0.75 units

So, any ideas how to do that - the line is obviously at a different angle to the rest so wont be as easy to calculate (and as you can see this CHGF normal vector has caused me problems and its a basis to this next calculation)