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- TL;DR Summary
- Skew symmetric form of cross product

I am sure you are all familiar with the cross product in 3D space.

Cyclic

Negative, if reversed, etc.

I am sure you are all familiar with the definition as: norm of the first vector, norm of the second, sine of the angle, perpendicular (but direction using right hand rule)

Sure, no problem.

And we can apply to the Moment = Distance cross Force.

And I can reason my way through why the geometric rule with the sine is justified, and, if the space is Euclidean, how we get the cyclic rule, above, too!

All fine.

But there is another way, as (I am sure) you all know.

Take the first vector in the cross product and use the components to form a skew symmetric matrix with positive determinant.

OK...

Has anyone ever "reasoned" their way as to why this skew symmetric matrix, times the column components of the second vector can give the same information as the traditional cross product?

I attach a picture, because I am not good with the equation writer here.

You see, I can reason out the trig-based operation (sine of the distance, magnitudes, etc, but I cannot reason out why this works... I can only say it is isomorphic to the traditional way of i cross j gives k.

**i**cross into**j**gives**k**.Cyclic

Negative, if reversed, etc.

I am sure you are all familiar with the definition as: norm of the first vector, norm of the second, sine of the angle, perpendicular (but direction using right hand rule)

Sure, no problem.

And we can apply to the Moment = Distance cross Force.

And I can reason my way through why the geometric rule with the sine is justified, and, if the space is Euclidean, how we get the cyclic rule, above, too!

All fine.

But there is another way, as (I am sure) you all know.

Take the first vector in the cross product and use the components to form a skew symmetric matrix with positive determinant.

OK...

Has anyone ever "reasoned" their way as to why this skew symmetric matrix, times the column components of the second vector can give the same information as the traditional cross product?

I attach a picture, because I am not good with the equation writer here.

You see, I can reason out the trig-based operation (sine of the distance, magnitudes, etc, but I cannot reason out why this works... I can only say it is isomorphic to the traditional way of i cross j gives k.