# Problem Understanding theorm of Cross Product

## 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
Mentor

## 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 Helper

## 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.

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Mark44
Mentor
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
Homework Helper
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
Staff Emeritus
Gold Member
2021 Award
Dick, my mind is blown right now.

Dick
Homework Helper
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.

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D H
Staff Emeritus