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## 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?

- Thread starter PsychonautQQ
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- #1

- 784

- 11

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?

- #2

Mark44

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

- #3

Dick

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

Last edited:

- #4

Mark44

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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)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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Dick

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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).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.

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Office_Shredder

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Dick, my mind is blown right now.

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Dick

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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.Dick, my mind is blown right now.

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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.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.

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