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anonymous
 one year ago
My question is with respect to PS 12b. 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.
anonymous
 one year ago
My question is with respect to PS 12b. 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
 one year ago
Best ResponseYou've already chosen the best response.1Can you make a screen shot of the question/answer and post it?

phi
 one year ago
Best ResponseYou've already chosen the best response.1It 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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