If (a,b)(a,b) is a are the coordinates of a point in Cartesian Plane, uu is its magnitude and alphaα is its angle then (a,b)(a,b) in Polar Form is written as (u,alpha)(u,α).
Magnitude of a cartesian coordinates (a,b)(a,b) is given bysqrt(a^2+b^2)√a2+b2 and its angle is given by tan^-1(b/a)tan−1(ba)
Let rr be the magnitude of (3sqrt3,-3)(3√3,−3) and thetaθ be its angle.
Magnitude of (3sqrt3,-3)=sqrt((3sqrt3)^2+(-3)^2)=sqrt(27+9)=sqrt36=6=r(3√3,−3)=√(3√3)2+(−3)2=√27+9=√36=6=r
Angle of (3sqrt3,-3)=Tan^-1((-3)/(3sqrt3))=Tan^-1(-1/sqrt3)=-pi/6(3√3,−3)=tan−1(−33√3)=tan−1(−1√3)=−π6
implies⇒ Angle of (3sqrt3,-3)=-pi/6(3√3,−3)=−π6
This is the angle in clockwise direction.
But since the point is in fourth quadrant so we have to add 2pi2π which will give us the angle in anti-clockwise direction.
implies⇒ Angle of (3sqrt3,-3)=-pi/6+2pi=(-pi+12pi)/6=(11pi)/6(3√3,−3)=−π6+2π=−π+12π6=11π6
implies⇒ Angle of (3sqrt3,-3)=(11pi)/6=theta(3√3,−3)=11π6=θ
implies (3sqrt3,-3)=(r,theta)=(6,(11pi)/6)⇒(3√3,−3)=(r,θ)=(6,11π6)
implies (3sqrt3,-3)=(6,(11pi)/6)⇒(3√3,−3)=(6,11π6)
Note that the angle is given in radian measure.
Also the answer (3sqrt3,-3)=(6,-pi/6)(3√3,−3)=(6,−π6) is also correct.
