Let length of vector= r=√(ax2+ay2)
Let θ be the angle with x-axis.
ax=rCosθ
ay=rSinθ
a'x =rCos(θ-α)= r(CosθCosα+SinθSinα)
= axCosα+aySinα
a'y = rSin(θ-α)= r(SinθCosα-CosθSinα)
= ayCosα - axSinα
2 Answers
metal
·2009-06-07 06:10:39
kamalendu ghosh
·2009-06-07 06:10:50
let the initial angle of poinr P be θ....
let r be the dist of point from origin
then in new system
X=rcos(θ-α)
and in the old sys x=rcosθ
similary
Y=rsin(θ-α)
..
y=rsinθ
now substitute the values....
as tanθ=ay/ax