Cross product

  • #26
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No, that cross product is not zero. The dot product of u with it is zero.
If you have a triple product a.(b x c) and a is parallel to either b or c then the triple product is zero. The reason is clear: b x c is perpendicular to b and c; if a is parallel to b then a is perpendicular to b x c.
since the rE x W and rb xTB are not perpendicular to u , so it,s not = 0 ?
how do u knw that a is parallel to b then a is perpendicular to b x c ??
 
  • #27
haruspex
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how do u knw that a is parallel to b then a is perpendicular to b x c ??
The cross product of two vectors (if nonzero) is always perpendicular to the two vectors.
If a is parallel to b then it must be perpendicular to b x c.
 
  • #28
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The cross product of two vectors (if nonzero) is always perpendicular to the two vectors.
If a is parallel to b then it must be perpendicular to b x c.
how do u knw that the rE x W and rb xTB are not perpendicular to u ????
 
  • #29
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how do u knw that the rE x W and rb xTB are not perpendicular to u ????
Why would that matter? Nothing in the algebra assumes those are not zero.
 
  • #30
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Why would that matter? Nothing in the algebra assumes those are not zero.
because when rE x W and rb xTB are not perpendicular to u , then i cant say that u . (rE x W) and u .(rb xTB ) = 0
 
  • #31
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because when rE x W and rb xTB are not perpendicular to u , then i cant say that u . (rE x W) and u .(rb xTB ) = 0
Nobody is saying those two dot products are individually zero.
The balance of torques gives us that rE x W + rb x TB + rDA x force_at_A = 0.
We take the dot product of that with u to obtain the equation u.(rE x W) + u.(rb x TB) + u.(rDA x force_at_A) = 0.
We then observe that since u is parallel to rDA the triple product u.(rDA x force_at_A) must be zero.
From that we conclude u.(rE x W) + u.(rb x TB) = 0.
This does not tell us anything about whether u is parallel to any of the four other vectors in that equation.
 

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