
#1
Sep811, 08:30 PM

P: 200

1. The problem statement, all variables and given/known data
Using these 2 vectors: [itex] \vec u = (3,4,0)[/itex] [itex] \vec v = (1,1,1)[/itex] I must verify that theta is the same with these 2 equations: Dot product [itex] \vec u \bullet \vec v = \vec u \vec v cos( \theta)[/itex] Cross product [itex] \vec u \wedge \vec v = \vec u \vec v sin( \theta)[/itex] 2. Relevant equations They were given in 1) 3. The attempt at a solution I did all the calculations, I get the following answers: [itex] \vec u  = 5[/itex] [itex] \vec v  = \sqrt{3}[/itex] [itex] \vec u \bullet \vec v = 1[/itex] [itex] \vec u \wedge \vec v =\sqrt{74}[/itex] I then solve the 2 equations given above using arcsin and arccos to find the values of theta, but I get 96.6 using the dot product, and 83.3 using the cross product. The weird thing is that 18083.3 = 96.6... I must be missing something obvious, but I can't understand why I get the wrong answer :( 



#2
Sep811, 08:48 PM

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P: 3,309

2 vectors define a plane
note that within that plane you can consider the smaller (<=90) or larger (>=90) angle between the 2 vectors , but they will always sum to 180degrees 



#3
Sep811, 09:01 PM

P: 200

How can you tell which value is the "right" one when trying to determinate the angle between the 2 vectors using the cross product? I'm trying to visualize the vectors in my head, and I know there is only one "right" answer. http://i.imgur.com/uM2ni.jpg 



#4
Sep811, 09:13 PM

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Thanks
PF Gold
P: 11,534

Angle between 2 vectors using 1) Dot product and 2) cross product gives diff. answer?
Hint: Arcsin will always give you a result between 90 degrees and 90 degrees, yet the angle between two vectors ranges from 0 to 180 degrees.




#5
Sep811, 09:16 PM

P: 200





#6
Sep811, 09:21 PM

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P: 3,309

note that the cross product expression is a magnitude
[tex] u \times v = uvsin(\theta) \geq 0 [/tex] the dot product allows negative values which will occur when the angel is greater than 90 degrees so in short, use the dot product 



#7
Sep811, 09:22 PM

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You can't conclusively determine the angle from the arcsin alone, just as you can't tell me what x equals with certainty if all I told you is sin x = 0.5.




#8
Sep811, 10:10 PM

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P: 3,309

To add onto vela's comments
Consider a plot of sin(t) with t from pi to pi In the region pi to 0 , sin(t) is negative. As you are dealing with magnitudes [itex] \frac{ u \times v}{ uv}[/itex] will never be negative, so the arcsin will only return a value in the range 0 to pi now on a plot of 0 to pi, the graph of sin(t) is symmetric about pi/2. So say you know sin(t) = 0.5. This could be either t=30 or t=150. The calculator will always return a number in the range (90 to 90) so in this case 30deg. 



#9
Sep911, 10:19 PM

P: 200

Ok I get it, thanks guys :D



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