Find the position vector of a point A such that OA is inclined to OX at 60 degrees and to OY at 45 degrees and ||OA|| = 10 units
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You can use the cosines of the vector to determine the answer and then just scale it by 10, x/L = cos(\alpha)=cos(60) x=L*sqrt(3)/2 y/L = cos(\beta)=cos(45) y=L*1/sqrt(2) and L = 10, OA = 10*sqrt(3)/2 i + 10*1/sqrt(2) j
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