# Homework Help: Proof this cross product property

1. Jan 26, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data
Show that :

|| a x b||^2 = ||a||^2||b||^2 - (a . b)^2

2. Relevant equations

3. The attempt at a solution

I don't know where to start off

Last edited: Jan 26, 2009
2. Jan 26, 2009

### Dick

What's the difference between ||axb||^2 and (axb)^2? Don't you mean (a.b)^2 instead of (axb)^2? Just use expressions for the magnitude of the cross and dot product and the angle between the vectors a and b.

3. Jan 26, 2009

### Staff: Mentor

What you're supposed to show doesn't make sense.
Given that, I understand what is meant by || a x b||^2, ||a||^2, and ||b||^2, but I have no idea what (a x b)^2 even means, let alone how to compute it.

Can you clarify what is meant here by (a x b)^2?

4. Jan 26, 2009

### -EquinoX-

sorry a typo, I've revised the question, it should be (a.b)^2

I know that a x b = ||a|| ||b|| sin x, so then how do I find the magnitude of this? I am confused because there's the sin function...

5. Jan 26, 2009

### Dick

Then express both side in terms of |a|, |b| and the angle between them.

6. Jan 26, 2009

### -EquinoX-

not sure I understand what you mean

7. Jan 26, 2009

### Dick

If t is the angle between a and b, then a.b=|a|*|b|*cos(t). There a similar relation for |axb|.

8. Jan 26, 2009

### -EquinoX-

ok is this correct?

$$|a\times b|^2 = |a|^2|b|^2\sin^2\theta = |a|^2|b|^2(1 - \cos^2\theta)$$
$$= |a|^2|b|^2 - |a|^2|b|^2\cos^2\theta = |a|^2|b|^2 - (a\cdot b)^2$$

9. Jan 26, 2009

### Dick

Absolutely.

10. Jan 26, 2009

### -EquinoX-

I am not sure of the notations through can I write |a x b|^2 as ||a x b||^2 as it denotes magnitude?

11. Jan 26, 2009

### Dick

They both mean the same thing to me. If you're a stickler you might want to demand writing ||v|| if v is a vector and |x| if x is a number.

12. Jan 26, 2009

### -EquinoX-

well yes, but what's meant here by ||v|| here is the magnitude of vector v right? now I am confused though, by definition of the geometric, it says that:

a x b = ||a|| ||b|| sin x

and it's not

|a x b| = ||a|| ||b|| sin x

can you explain this?

13. Jan 26, 2009

### Staff: Mentor

I don't understand what you're saying above. Does "and it's not" refer to the first equation or the second? Either way, the first equation is incorrect and the second one is correct (aside from inconsistent usage of | | and || ||).

a X b is a vector, which ||a|| ||b|| sin x is the product of three numbers, and so is a number.

14. Apr 25, 2009

### natangwe

Can you please prove for me this identity: A x (B x C) = (A.C)B - (A.B)C

15. Apr 25, 2009

### natangwe

Can you please prove for me this identity, analytically: A x (B x C) = (A.C)B - (A.B)C and llU x Vll^2 = llUll^2 llVll^2 - (U.V)^2. Any one with understanding on this please help out.