 #1
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All necessary information is attached except the answer in Cartesian coordinates, which is ixjy+2kz and my work converting back from cylindrical to Cartesian, which I used WolframAlpha for, as the trig is a mess (that is, if the way I am doing this is correct).
http://www.wolframalpha.com/input/?i=%28icos%28tan^1%28y%2Fx%29%29%2Bjsin%28tan^1%28y%2Fx%29%29%29%28sqrt%28x^2%2By^2%29%28cos^2%28tan^1%28y%2Fx%29%29sin^2%28tan^1%28y%2Fx%29%29%29%29%2B%28isin%28tan^1%28y%2Fx%29%29%2Bjcos%28tan^1%28y%2Fx%29%29%29%282z%28cos^2%28tan^1%28y%2Fx%29%29sin^2%28tan^1%28y%2Fx%29%29%29%29&a=i_Variable
WolframAlpha, however, just gave me this mess back. Would appreciate if someone would let me know if the way I'm going about this is correct. I have a strong feeling the matrix I'm using for reconversion is off.
http://www.wolframalpha.com/input/?i=%28icos%28tan^1%28y%2Fx%29%29%2Bjsin%28tan^1%28y%2Fx%29%29%29%28sqrt%28x^2%2By^2%29%28cos^2%28tan^1%28y%2Fx%29%29sin^2%28tan^1%28y%2Fx%29%29%29%29%2B%28isin%28tan^1%28y%2Fx%29%29%2Bjcos%28tan^1%28y%2Fx%29%29%29%282z%28cos^2%28tan^1%28y%2Fx%29%29sin^2%28tan^1%28y%2Fx%29%29%29%29&a=i_Variable
WolframAlpha, however, just gave me this mess back. Would appreciate if someone would let me know if the way I'm going about this is correct. I have a strong feeling the matrix I'm using for reconversion is off.
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