# Questions on Cross Product

1. Jan 17, 2013

### doctorjuice

First question:

a x b = -b x a
Why is this so?

As I understand, a major purpose of the cross product (if not, the purpose) is to find a third vector that is perpendicular to two other vectors simultaneously. Let's say a x b = c. Shouldn't the answer really be, a x b = +/- c? Since, of course, both c and -c are perpendicular to a and b simultaneously.

The situation of the sqrt(4) = +/- 2 is analogous to this.

Second question:

The cross product is said to be something that only works in 3 dimensional space. In 2D, it is said to be not applicable. Take two vectors in 2D that are just the negative of each other. Obviously a line can be drawn straight up relative to these two vectors that is perpendicular to both (visualize an upside down T). Since the cross product's main purpose is to find a third vector that is perpendicular to two other vectors, can it be said that the cross product fails in 2 dimensional space?

Thanks for your time.

2. Jan 17, 2013

### Fredrik

Staff Emeritus
This follows easily from the definition. What definition have you been taught?

You may have to define the word "should".

Torque is a good example of the use of the cross product in physics. When you use a screwdriver on a screw, you're applying torque that's equal to a cross product. The direction of that vector contains the information about which way you're turning the screw. Angular momentum is another good example. In this case, the direction of the vector corresponds to the direction of the rotation. So that direction isn't irrelevant.

You may also have to define the word "fail".

I can't say that the cross product fails in 2-dimensional space, since there is no cross product on $\mathbb R^2$. But you are right that there are no vectors that are perpendicular to two linearly independent vectors.

3. Jan 17, 2013

### lurflurf

0=(a+b)x(a+b)=a x a+a x b+b x a+b x b=a x b+b x a
if
a x b = c
t c is perpendicular to both a and b for any t
The cross product is chosen to preserve distance and to be right handed.

There are generalizations of the cross product to other spaces, none of them have all the properties of the n=3 case, one reason is that several function are the same only for this case
for example consider functions
$$V \times V \rightarrow V \\ V \times V \rightarrow V^{n-2} \\ V \times V^{n-2} \rightarrow V \\$$
only when n=3 can we hope these functions are the same

There is the http://en.wikipedia.org/wiki/Seven-dimensional[/PLAIN] [Broken] cross_product]Seven dimensional cross product which is the closest.

Last edited by a moderator: May 6, 2017
4. Jan 17, 2013

### micromass

The definition is that the square root is always nonnegative. Thus by definition, we have $\sqrt{4}=2$. We do not define $\sqrt{4}=-2$ or $\sqrt{4}=\pm 2$. We only give one value to the square root because we want it to be a function.

5. Jan 17, 2013

### CompuChip

Similarly, for the cross product, the definition is such that the cross product is "right-handed". That is, if you take a right handed coordinate system (turning the fingers of your right hand from the x to the y-axis your thumb points along the z-axis) then ex x ey = ez (or, in alternative notation, $\hat i \times \hat j = \hat k$) rather than -ez.

6. Jan 17, 2013

### doctorjuice

I've been taught the same definition, I was just asking why it was defined this way, I didn't see the reason for it at first. There are actually a couple reasons, I believe (and I would like to hear confirmation or refutal of these reasons):

1) If a x b = +/-c (instead of +c), then the expression a x b would be ambiguous. By itself, this is not a major problem but once complicated calculations with cross products occur this problem becomes very large.

Analogous to sqrt(4), when it is part of complex calculations, if it were to ambiguously mean +/-2, then the whole mathematical expression of a complex operation involving multiple sqrt(4) terms would be very ambiguous and you could come up with many different answers for the same problem.

2) Even if the sign of c does not affect the end result in mathematics, in physics it does, and the information is very important in physics (as you said in other parts of your post).

I think you misunderstood this part of my post. I was referring to two vectors in 2D, that go in opposite directions (so a and -a would be a pair of such vectors). If a vector is drawn straight up in 2D relative to these two vectors (think an upside-down T, with the left part being vector a, right part being vector b (or -a) and the up part being a vector perpendicular to both, vector c) then that vector is perpendicular to both a and b in 2D space simultaneously.

Since a major purpose of the cross-product is to find a vector that is perpendicular to two vectors simultaneously, I was making the assertion that the cross product fails at fulfilling its purpose in 2D space, for this specific situation.

I would love to hear yours and anybody else's thoughts on what I have said here.

7. Jan 17, 2013

### CompuChip

1) is definitely true. It's just a matter of convention, and just like we chose sqrt(4) to give 2 and not -2 because that's often easier, we also choose (1,0,0) x (0,1,0) to be (0,0,1) and not (0,0,-1) because it's usually easier. You could view the definition of sqrt as just some function, which happens to have the property that it squares to its argument without claiming that there are no other such numbers, you could see the definition of the cross product in a similar way.

I'm not sure about 2) - I think quite the opposite is true. It does not matter which way you define it, but if you get it the other way around you might also want to change other conventions so you don't end up writing minus signs all the time.

8. Jan 17, 2013

### THSMathWhiz

A formal cross product is defined for $\mathbb{R}^3$ and $\mathbb{R}^7$. There is a projective cross product for $\mathbb{R}^2$, which returns a directed scalar. Given $\mathbf{u}=\langle u_1,u_2\rangle$ and $\mathbf{v}=\langle v_1,v_2\rangle$, the "cross product" is $\mathbf{u}\times_2\mathbf{v}=(\langle u_1,u_2,0\rangle\times_3\langle v_1,v_2,0\rangle)\cdot\langle0,0,1\rangle$, where $\times_n$ is the cross product in $\mathbb{R}^n$.

9. Jan 17, 2013

### Fredrik

Staff Emeritus
Sorry about that. I just saw that you were asking about cross products in 2D and jumped to the wrong conclusion about what you wanted to ask. I guess I should have actually read the question.

You could certainly define a "product" of two linearly dependent vectors in ℝ2 that has a vector perpendicular to both as the result. For example, you could define
$$(a_1,a_2)(b_1,b_2)=\operatorname{sgn}(a_1b_1)(a_2,-a_1)$$ where sgn is the sign function. But this seems rather pointless. This product also has very little in common with the cross product on ℝ3.

10. Jan 18, 2013

### lurflurf

1&2)The convention does not matter, but using multiple conventions would lead to errors. This is the same in mathematics and physics.

As for finding a vector that is perpendicular, as I mentioned above the cross product serves several purposes simultaneously in 3-space this cannot be done in other spaces. One generalization gives the (unique up to scalar multiplication) vector perpendicular to n-1 given vectors. In 2-space this means we take one vector and return a vector perpendicular to it. In 11-space this means we take ten vectors and return a vector perpendicular to it. Only in 3-space can we take two vectors and return one vector perpendicular to it.