# Homework Help: Vector proofs

1. Jun 10, 2006

### masterofthewave124

i have to prove the diagonals of a rhombus intersect at right-angles using the scalar dot product.

i have set up a cartesian plane system where A lies on the origin. numbering the sequential points clockwise, i let B = (a,b), C = (a+c, b) and D = (c,0). I then thought if i set up vectors AC and DB and found the dot product between them, i could get a value of zero. unfortunately thats not the case so where did i go wrong?

2. Jun 10, 2006

### Hootenanny

Staff Emeritus

3. Jun 10, 2006

### Hurkyl

Staff Emeritus
Well, he didn't set up a rhombus; he set up a parallelogram.

4. Jun 10, 2006

### Hootenanny

Staff Emeritus
So that would explain why my dot product was non zero. I fixed it for a rhombus and the proof works fine.

5. Jun 10, 2006

### masterofthewave124

hmm...where would my flaw lie? i think i just may be overlooking a critical property of a rhombus.

6. Jun 10, 2006

### Hurkyl

Staff Emeritus
What's the definition of a rhombus?

7. Jun 10, 2006

### masterofthewave124

a parallelogram having equal sides. i still can't see how the way i defined the coordinates is wrong, maybe just a little obscure (can be simplified).

8. Jun 10, 2006

### Hurkyl

Staff Emeritus
So did you, anywhere in your set-up, impose the condition that all the sides were equal?

9. Jun 10, 2006

### masterofthewave124

ahh im confused, the way i have my diagram, a traditional rhombus, it looks right. i can see how a square would be different.

so back to basics:
AB = BC = CD = DA

but that doesn't necessarily mean the coordinates will be represented that way, just the magnitude.

10. Jun 10, 2006

### Hurkyl

Staff Emeritus
What if you set a = 400, b = 400, and c = 1?

In the way you set up your problem, the vector AB = (a, b), and the vector AD = (c, 0). But clearly they are not automatically equal!

11. Jun 10, 2006

### masterofthewave124

hmm i see your point. but then what would AB = ?

basically a^2 + b^2 = c^2

so AB would equal (sqrt(c^2-b^2), sqrt (c^2 - a^2)), can't be right, much too messy...

12. Jun 10, 2006

### Hurkyl

Staff Emeritus
I suspect that this is all you need -- go back and redo your work, but make use of this relation.

Incidentally, this problem is easier if you don't use coordinates at all!

13. Jun 11, 2006

### masterofthewave124

oh ok, i just left my coordinates as is but when i got down to the dot product,

after i get a^2 + ac - ac - c^2 + b^2
i stated that a^2 + b^2 = c^2
so c^2 - c^2 = 0

i hope thats sufficient. what other method would have been easier Hurkyl?

14. Jun 11, 2006

### Hurkyl

Staff Emeritus
Do essentially the exact same thing, but never use coordinates: leave everything in terms of vectors.

15. Jun 11, 2006

### Hurkyl

Staff Emeritus
All right, I'll spoil the answer if you haven't worked it out yet yourself!

Let x and y be the vectors denoting two sides of our rhombus.

Then, as you noted, the two diagonals are:

x + y

and

x - y

and because the sides are equal:

x² = y²

The dot product of the diagonals is:

(x + y).(x - y) = x² - y² = 0

and thus, they are perpendicular.

(Where I've used the notation that x² means x.x)