Problem Understanding theorm of Cross Product

[PsychonautQQ]

Homework Statement

My textbook says:
The length of the cross product a x b is equal to the area of the parallelogram determined by a and b.

How can a length equal and area? They have different units?

 

[Mark44]

Homework Statement

My textbook says:
The length of the cross product a x b is equal to the area of the parallelogram determined by a and b.

How can a length equal and area? They have different units?

They are numerically equal. That doesn't say anything about units, just that the numbers are the same.

 

[Dick]

Homework Statement

My textbook says:
The length of the cross product a x b is equal to the area of the parallelogram determined by a and b.

How can a length equal and area? They have different units?

If a and b have dimension length, then axb properly has dimension (length)^2. So does the area of a parallelogram.

 

[Mark44]

If a and b have dimension length, then axb properly has dimension (length)^2.

You're going to have to convince me, Dick. a X b is a vector, and its magnitude is a length. I don't see how you can get (length)² out of a vector.

So does the area of a parallelogram.

 

[Dick]

You're going to have to convince me, Dick. a X b is a vector, and its magnitude is a length. I don't see how you can get (length)² out of a vector.

I'll try. The components of axb are products of components of a and b. If the components of a and b have dimension length, that means the components of axb have dimension (length)^2. Not all vectors have dimension length. An example from physics is that angular momentum is defined by l=rxp. r has dimensions of length (m), p has the dimensions of momentum (kg*m/s). l has dimensions of angular momentum (kg*m^2/s).

 

[Office_Shredder]

Dick, my mind is blown right now.

 

[Dick]

Dick, my mind is blown right now.

I'm kind of surprised this is of mind-blowing proportions. It's just consistently tracking units. In math, you largely treat vectors as dimensionless, so the issue never comes up. If your vectors are dimensionful and you want a length vector perpendicular to a and b, you should use axb/(length units). Which you usually do anyway, just by ignoring the (length)^2 aspect of axb. It should really be satisfying that |axb| correctly has the dimensions of a parallelogram area. Don't get me talking about differential forms and duality. It's why there is a 'cross product' only in three dimensions. The dimension funnyness reflects that.

 

[D H]

You're going to have to convince me, Dick. a X b is a vector, and its magnitude is a length. I don't see how you can get (length)² out of a vector.

I never use "length" as a synonym for the magnitude of a physical vector. What is the length of a velocity vector? That doesn't quite make sense. Velocity has dimensionality of length per time, not length. The magnitude of that velocity vector? That makes sense, and it even has a name: Speed.