# How to find the magnitude of this bivector?

1. Nov 29, 2011

### dimension10

How does one find the value of

$$||A \left(\hat{\imath}\wedge\hat{\jmath}\right) +B(\hat{\imath}\wedge\hat{k})+C( \hat{\jmath}\wedge\hat{k})||$$

Where A, B and C are constant coefficients.

2. Nov 29, 2011

### I like Serena

3. Nov 29, 2011

### micromass

It could also be that you actually did mean wedges and that you're working with the exterior algebra. In that case, you can define an inner product as follows

$$\langle v\wedge w,v^\prime\wedge w^\prime\rangle= det\left(\begin{array}{cc} \langle v,v^\prime\rangle & \langle v,w^\prime\rangle\\ \langle w,v^\prime\rangle & \langle w,w^\prime\rangle \end{array}\right)$$

Under this inner product, can you now calculate the norm in your example?

See http://en.wikipedia.org/wiki/Exterior_algebra for more details.

4. Nov 29, 2011

### dimension10

Thanks. However, I don't see any way to express the bivector in that way.

Last edited: Nov 29, 2011
5. Nov 29, 2011

### I like Serena

Ah, okay.
So it's a generalization of the cross product.

Well, it says here:
http://en.wikipedia.org/wiki/Exterior_algebra#Inner_product
that the 3 wedge products you have in your problem form an orthonormal basis.

And I take it the norm would be defined to be the square root of the bivector with itself?

6. Nov 29, 2011

### dimension10

But if I were to do that, then I would obtain a 2 by 3 matrix to take the determinant of and determinants are not defined for non-square matrices...

7. Nov 29, 2011

### I like Serena

How do you get a 2 by 3 matrix?

Note that "orthonormal basis" says it all... no need to take any determinants...

8. Nov 29, 2011

### dimension10

I don't exactly get how it is an orthonormal basis since A, B, and C are unknown coefficients.

If it were an orthonormal basis, then the magnitude would be 1, but I don't think this will be an orthonormal basis.

9. Nov 29, 2011

### I like Serena

Yes, A, B, and C are unknown scalar coefficients.

And:
$$\langle \left(\hat{\imath}\wedge\hat{\jmath}\right), \left(\hat{\imath}\wedge\hat{\jmath}\right) \rangle = 1$$$$\langle \left(\hat{\imath}\wedge\hat{\jmath}\right), (\hat{\imath}\wedge\hat{k}) \rangle = 0$$

10. Nov 30, 2011

### dimension10

But to get the bivector into that form, I would need to get the scalars outside the magnitude signs somehow...

11. Nov 30, 2011

### I like Serena

Perhaps you can use the following?
$$||x||=\sqrt{\langle x, x \rangle}$$$$\langle ax+by, z \rangle = a\langle x, z \rangle + b\langle y, z \rangle$$

12. Dec 1, 2011

### dimension10

Oh, so that is true for bivectors too? I thought it was true only for vectors! Thanks.

13. Dec 1, 2011

### I like Serena

The first is the "usual" definition of the norm.
Since you did not specify you were using another one (and there are many others), this one is assumed.

The second is part of the definition of the inner product, which is the generalized version of the dot product.