What are the components of the vector between the origin and the polar coordinate #(4, (11pi)/12)#?

1 Answer
Sep 30, 2017

The vector is #=-(sqrt2+sqrt6)hati+(sqrt6-sqrt2)hatj#

Explanation:

If the polar coordinates of the vector are #(r,theta)#, then

The components are #(rcostheta,rsintheta)# in the rectangular cordinates are

#x=rcosthetahati#

#y=rsinthetahatj#

Here, we have

#(r,theta)=(4,11/12pi)#

#x=4cos(11/12pi)=4cos(2/3pi+1/4pi)#

#=4(cos(2/3pi)cos(1/4pi)-sin(2/3pi)sin(1/4pi))#

#=4(-1/2*sqrt2/2-sqrt3/2*sqrt2/2)#

#=-4/4(sqrt2+sqrt6)#

#=-(sqrt2+sqrt6)#

#y=4sin(11/12pi)=4sin(2/3pi+1/4pi)#

#=4(sin(2/3pi)*cos(1/4pi)+cos(2/3pi)*sin(1/4pi))#

#=4*(sqrt3/2*sqrt2/2-1/2*sqrt2/2)#

#=(sqrt6-sqrt2)#

The vector is

#=-(sqrt2+sqrt6)hati+(sqrt6-sqrt2)hatj#