• Nov 24th 2008, 03:46 PM
AshleyT
I think its simple...i think i just have a brain block lol...mayb cuz its 12 at night.
OABC is a rectangle.
Quote:

A has position vector (2, -3, 5) (with respect to origin O)
B has position vector (x, 3, 7)(with respect to origin O)

So i thought, these must be perpendicular! Scale product...well...i got s = 13 which is wrong :(.

But also if these are perpendicular...won't they share the same x-coordinate? Or Y co-ordinate?

• Nov 24th 2008, 04:19 PM
Plato
Rectangles have right alngles.
So $\overrightarrow {OA} \cdot \overrightarrow {AB} = 0$.
• Nov 25th 2008, 08:55 AM
David24
Quote:

Originally Posted by AshleyT
I think its simple...i think i just have a brain block lol...mayb cuz its 12 at night.
OABC is a rectangle.

So i thought, these must be perpendicular! Scale product...well...i got s = 13 which is wrong :(.

But also if these are perpendicular...won't they share the same x-coordinate? Or Y co-ordinate?

hey mate,
As stated by Plato <OA,AB> = 0 as the shape is rectangular, have you been using <OA,OB> instead?
either way to solve AB, use AB = B - A

This will yield the correct solution,

Regards,

David
• Nov 25th 2008, 11:35 AM
Andreas Goebel
Hi,

I´d like to add an explanation why the vectors do not need to share a co-ordinate to be perpendicular.

Just look at the attached image.

Regards,

Andreas
• Nov 25th 2008, 11:39 AM
AshleyT
Aww, thank-you everyone for your help. I really appreciate it.