In this post, I want to summarize more about the cross product. We will first prove a simple fact about dot product.
It is well known that $ \vec{a} \cdot \vec{b} = |a| |b| \cos \theta $ where $ \theta $ is the angle between the two vectors, but why?
To see that, note that the dot product can be written as a matrix product $ x^Ty $ where vectors are written as column. Now we insert $ R^TR = I $ into the product with $ R $ being a rotation (i.e. orthogonal), we have $ x^Ty = x^TR^TRy = (Rx)^TRy $, that mean dot product is rotational invariant.
Now we scale all vectors to have unit length and rotate them such that they both lie on the same XY plane with $ \vec{x} $ being parallel to the $ x $ axis, it is obvious that it is possible, then the rest is obvious. The dot product is simply the length of the x projection of the other vector, which is exactly $ \cos \theta $! Mapping it back to the original problem, multiplying back the scale, we get the formula we want.
The next interesting thing is to prove the also familar identity $ \vec{a} \times \vec{b} = |a||b|\sin \theta $. To that end, it is easier to prove $ |\vec{a} \times \vec{b}|^2 = |a||b| - (\vec{a} \cdot \vec{b})^2 $, as this will automatically implies the result. To prove this one, however, is easy too as one only need to multiply out all the terms, detail skipped.
It is well known that $ \vec{a} \cdot \vec{b} = |a| |b| \cos \theta $ where $ \theta $ is the angle between the two vectors, but why?
To see that, note that the dot product can be written as a matrix product $ x^Ty $ where vectors are written as column. Now we insert $ R^TR = I $ into the product with $ R $ being a rotation (i.e. orthogonal), we have $ x^Ty = x^TR^TRy = (Rx)^TRy $, that mean dot product is rotational invariant.
Now we scale all vectors to have unit length and rotate them such that they both lie on the same XY plane with $ \vec{x} $ being parallel to the $ x $ axis, it is obvious that it is possible, then the rest is obvious. The dot product is simply the length of the x projection of the other vector, which is exactly $ \cos \theta $! Mapping it back to the original problem, multiplying back the scale, we get the formula we want.
The next interesting thing is to prove the also familar identity $ \vec{a} \times \vec{b} = |a||b|\sin \theta $. To that end, it is easier to prove $ |\vec{a} \times \vec{b}|^2 = |a||b| - (\vec{a} \cdot \vec{b})^2 $, as this will automatically implies the result. To prove this one, however, is easy too as one only need to multiply out all the terms, detail skipped.
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