# Cross product vectors problem

1. Aug 6, 2013

### phydis

a and b are two vectors and x is the angle between them.

||axb|| = ||a||||b||sinx ------(1)
||axb|| = ||a||||b||||sinx|| ------(2)

which one is correct? why?

2. Aug 6, 2013

### SteamKing

Staff Emeritus
1. That is one definition of cross product.

3. Aug 6, 2013

### CAF123

x is the smallest angle between vectors a and b. You should convince yourself that this implies both forms are equivalent.

4. Aug 6, 2013

### phydis

if x is the smallest angle between a and b, when n is the unit vector perpendicular to both a and b,

axb = ||axb||n

am i correct?

5. Aug 6, 2013

### Zondrina

Your first one is correct :

$||axb|| = ||a|| ||b|| sin(θ)$

6. Aug 6, 2013

### CAF123

Yes

I think the OP also wanted to know why. I would say the two forms are equivalent. What is your argument for why the other form is incorrect?

7. Aug 6, 2013

### phydis

if two forms are equivalent then ||sinx||= sinx (where x is the smallest angle between a and b)
how can you explain this?

8. Aug 6, 2013

### CAF123

Given that x is the smallest angle between a and b what is the possible values of x?

9. Aug 6, 2013

### lurflurf

This has to do with what values of x we allow. We need the absolute value if we are to allow x such that sin x<0

10. Aug 6, 2013

### phydis

assume x= 11∏/6 , then smallest angle between a and b ∏/6, but when applying it to x it should be -∏/6. shouldn't it? (if it shouldn't then why?)

are ||sin(-∏/6)|| and sin(-∏/6) equivalent?

11. Aug 6, 2013

### CAF123

No, x is restricted to lie within the interval $[0, \pi]$. Check your book. This condition will come with the definition of cross product.

12. Aug 6, 2013

### lurflurf

Who cares if x is in [0,pi) or not. We have |sin x| if sin x is already positive sin x will do.

13. Aug 6, 2013

### CAF123

Yes, so if $x \in [0,\pi]$ then $|a \times b| = |a| |b| \sin x$ since sin x is positive. If x is not in this interval, then $|a \times b| = |a| |b| |\sin x|$ (the magnitude of $a\times b$ is not negative so we take the modulus of sin x)

In any book I have read, they put the condition that x is in [0,π], so you see it written $|a \times b| = |a| |b| \sin x$ more commonly.

14. Aug 6, 2013

### phydis

According to the way of my thinking ||axb|| = ||a||||b||sinx is enough to represent magnitude of axb for all x angle.

let n be an unit vector perpendicular to both a and b vectors. therefore we get,

||axb|| = || ||a||||b||sinx n ||

when x > 0, ||axb|| = ||a||||b||sinx ||n||
since ||n|| = 1 , ||axb|| = ||a||||b||sinx

when x<0, ||axb|| = || ||a||||b||sin(-x) n || = ||a||||b||sinx ||-n|| (with right hand rule)
since ||-n|| = 1 , ||axb|| = ||a||||b||sinx

although the other form (2) gives the same magnitude, it is theoretically incorrect.
and only this imaginary of mine is not confusing me. I wanna know does this really happen there?