What is the distance between $(- 3, \frac{17 π}{12})$ $(-3 , (17 pi)/12 )$ and $(- 1, \frac{π}{4})$ $(-1 , pi/4 )$ ?

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Answer 1 · 2016-07-15T00:39:38

The polar coordinates of two points.

$A \to r_{1} = - 3, θ_{1} = \frac{17 π}{12}$ $A->r_1=-3,theta_1=(17pi)/12$

$B \to r_{2} = - 1, θ_{2} = \frac{π}{4}$ $B->r_2=-1,theta_2=pi/4$

Then the distance between A,B

$d = \sqrt{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} cos (θ_{1} - θ_{2})}$ $d=sqrt(r_1^2+r_2^2+2r_1r_2cos(theta_1-theta_2))$

$d = \sqrt{{(- 3)}^{2} + {(- 1)}^{2} + 2 (- 3) (- 1) cos (\frac{17pi}{12} - \frac{π}{4})}$ $d=sqrt((-3)^2+(-1)^2+2(-3)(-1)cos("17pi"/12-pi/4))$

$\Rightarrow d = \sqrt{9 + 1 + 6 \cdot cos (\frac{14pi}{12})}$ $=>d=sqrt(9+1+6*cos("14pi"/12))$

$\Rightarrow d = \sqrt{9 + 1 + 6 \cdot cos (\frac{7pi}{6})}$ $=>d=sqrt(9+1+6*cos("7pi"/6))$

$\Rightarrow d = \sqrt{9 + 1 + 6 \cdot cos (π + \frac{π}{6})}$ $=>d=sqrt(9+1+6*cos(pi+pi/6))$

$\Rightarrow d = \sqrt{10 - 6 \cdot cos (\frac{π}{6})}$ $=>d=sqrt(10-6*cos(pi/6))$

$\Rightarrow d = \sqrt{10 - 6 \cdot \frac{\sqrt{3}}{2}}$ $=>d=sqrt(10-6*sqrt3/2)$

$\Rightarrow d = \sqrt{10 - 3 \sqrt{3}}$ $=>d=sqrt(10-3sqrt3)$

What is the distance between (−3,17π12)(-3 , (17 pi)/12 ) and (−1,π4)(-1 , pi/4 )?