What are the components of the vector between the origin and the polar coordinate (4, (11pi)/12)(4,11π12)?

1 Answer
Sep 30, 2017

The vector is =-(sqrt2+sqrt6)hati+(sqrt6-sqrt2)hatj=(2+6)ˆi+(62)ˆj

Explanation:

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

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

x=rcosthetahatix=rcosθˆi

y=rsinthetahatjy=rsinθˆj

Here, we have

(r,theta)=(4,11/12pi)(r,θ)=(4,1112π)

x=4cos(11/12pi)=4cos(2/3pi+1/4pi)x=4cos(1112π)=4cos(23π+14π)

=4(cos(2/3pi)cos(1/4pi)-sin(2/3pi)sin(1/4pi))=4(cos(23π)cos(14π)sin(23π)sin(14π))

=4(-1/2*sqrt2/2-sqrt3/2*sqrt2/2)=4(12223222)

=-4/4(sqrt2+sqrt6)=44(2+6)

=-(sqrt2+sqrt6)=(2+6)

y=4sin(11/12pi)=4sin(2/3pi+1/4pi)y=4sin(1112π)=4sin(23π+14π)

=4(sin(2/3pi)*cos(1/4pi)+cos(2/3pi)*sin(1/4pi))=4(sin(23π)cos(14π)+cos(23π)sin(14π))

=4*(sqrt3/2*sqrt2/2-1/2*sqrt2/2)=4(32221222)

=(sqrt6-sqrt2)=(62)

The vector is

=-(sqrt2+sqrt6)hati+(sqrt6-sqrt2)hatj=(2+6)ˆi+(62)ˆj