How do you find the distance between #(2, π/3), (5, 2π/3)#?

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Answer 1 · 2017-06-11T15:54:40

The distance is #=4.63#

We convert the polar coordinates #(r,theta)# into rectangular coordinates #(x,y)#

#(r,theta)#, #=>#, #(x,y)=(rcostheta, rsintheta)#

#(2,1/3pi)#, #=>#, #(2cos(1/3pi),2sin(1/3pi))=(1,sqrt3)#

and

#(5,2/3pi)#, #=>#, #(5cos(2/3pi),5sin(2/3pi))=(-5/2,5/2sqrt3)#

The distance is between #(1,sqrt3)# and #(-5/2,5/2sqrt3)#

#d=sqrt((-5/2-1)^2+(5/2sqrt3-sqrt3)^2)#

#=sqrt(49/4+27/4)#

#=8.72/2#

#=4.36#

Answer 2 · 2017-06-11T16:05:04

Here is a graph of vectors from the origin to the two points:
Desmos.com

Please observe that the line connecting the two points forms a triangle:
Desmos.com

We can use the Law of Cosines to find the length of the blue line (side c):

#c^2 = a^2 + b^2 -2(a)(b)cos(theta)#

Where #a = 5, b = 2, and theta = (2pi)/3-pi/3 = pi/3#

#c^2 = 5^2 + 2^2 -2(5)(2)cos(pi/3)#

#c^2 = 25+4 - 2(10)(1/2)#

#c^2 = 19#

#c = sqrt19 larr# answer