Answer
(a) See explanations.
(b) $2.91$
(c) $2.91$
Work Step by Step
(a) Form a triangle with the three points: the origin (pole) O, polar point P1 $(r_1, \theta_1)$, and polar point P2 $(r_2, \theta_2)$. The lengths of two sides are $r_1$, $r_2$ and the angle between them is $ \theta_2- \theta_1$. Use the Law of Cosines, the third side or the distance (d) between P1 and P2 is given by $d^2=r_1^2+r_2^2-2r_1r_2cos(\theta_2- \theta_1)$ which gives $d=\sqrt {r_1^2+r_2^2-2r_1r_2cos(\theta_2- \theta_1)}$
(b) Given P1 $(3, \frac{3\pi}{4})$, P2 $(1, \frac{7\pi}{6})$, plug-in the number in the equation above, we have
$d=\sqrt {3^2+1^2-2\times3\times1cos(\frac{7\pi}{6}-\frac{3\pi}{4})}\approx2.91$
(c) Covert from polar to rectangular coordinates using the formula $x=r\cdot cos\theta$ and $y=r\cdot sin\theta$, we get P1 $(3cos \frac{3\pi}{4}, 3sin \frac{3\pi}{4})$ and P2 $(cos \frac{7\pi}{6}, sin \frac{7\pi}{6})$. Plug-in the numbers in the distance formula between two points, we get $d'=\sqrt {(cos \frac{7\pi}{6}-3cos\frac{3\pi}{4} )^2+( sin\frac{7\pi}{6}-3sin \frac{3\pi}{4})^2}\approx2.91$ which is the same as the result from (b)
