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judge dreadz

maths - sin between vectors

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Quote:
Original post by WhardieJones
|| A x B || = ||A|| ||B|| sin(angle)

The formula is not correct.
A.B = |A||B| cos(angle)
A perpendicular to B gives A.B = 0 and sin(pi/2) = 1 AFAIK.

Together with sin*sin+cos*cos = 1 you end up with

A.B
sin(a) = sqrt(1 - (------)2)
|A||B|


'.' beeing the dot product, of course.

This is another solution to your problem. I guess you'll find many more (playing with sin, cos and tan is rather fun :)) be aware that you are limited to values of sin(a) that are in [0,1] - ie a in[0,pi]. In fact, the formula dont' allow you to know if you are in [0,pi] or in [pi,pi*2] (because of the sqrt)

Regards,

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Quote:
Original post by Emmanuel Deloget
Quote:
Original post by WhardieJones
|| A x B || = ||A|| ||B|| sin(angle)

The formula is not correct.
A.B = |A||B| cos(angle)
I think he means for x to be the cross product, in which case the formula is correct.

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Guest Anonymous Poster
Quote:

|| A x B || = ||A|| ||B|| sin(angle)


Quote:
Original post by Emmanuel Deloget
'.' beeing the dot product, of course.


'x' being the cross product, of course.

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Quote:
Original post by Anonymous Poster
Quote:

|| A x B || = ||A|| ||B|| sin(angle)


Quote:
Original post by Emmanuel Deloget
'.' beeing the dot product, of course.


'x' being the cross product, of course.


Thanks for the clarification, mister anonymous poster [razz]

Since it was not very clear that the OP needed to have a 3D formula, I incorrectly assumed that cross product wouldn't be used. Thus, in my mind, the answer was "dot product" and I thought the formula was wrong.

I praise you, dear AP, to allow me to apologize for this horrible, terrible, deadly mistake.

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Quote:
Original post by Emmanuel Deloget
I praise you, dear AP, to allow me to apologize for this horrible, terrible, deadly mistake.


In the name of Euclid, the Pythagoreans, and Euler, your sines are forgiven you. May Gauss have mercy on your soul!

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