My question is with respect to PS 1-2b. There appears to some inconsistency between the diagram and the stated solution. It seems that the answer should be the (dir(u)+dir(v))/|dir(u)+dir(v)|, where dir(x) means unit vector in the direction of x.

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My question is with respect to PS 1-2b. There appears to some inconsistency between the diagram and the stated solution. It seems that the answer should be the (dir(u)+dir(v))/|dir(u)+dir(v)|, where dir(x) means unit vector in the direction of x.

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  • phi
Can you make a screen shot of the question/answer and post it?
  • phi
It looks like they were sloppy writing up the answer. First, to find the vector that bisects u and v normalize both to the same length (unit length is fine) thus begin with \[ \hat {u} = \frac{\vec{u}}{|u|} \] and similarly for \(\hat v\) as noted in the answer, \( \hat u + \hat v\) bisects the angle between the two vectors. They want the unit length vector. to make this vector unit length , divide by its magnitude, we get \[ \frac{\hat u + \hat v}{| \hat u + \hat v| } \] this is what I assume you mean by (dir(u)+dir(v))/|dir(u)+dir(v)| notice that the answer starts by stating u and v are unit length and they then give the answer \[ \frac{u + v}{| u+ v|} \] which is the correct answer (if u and v are unit length) the problem is that u and v in the question are not unit length thus, in the answer, they should have started with \(\hat u \) and \(\hat v\) (i.e. u and v normalized to unit length). I would say they were a wee bit sloppy, but their answer is essentially correct. (they should put "hats" over u and v)

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Thanks again phi.

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