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# Help Pls

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Let $$A = (3, \theta_1)$$ and $$B = (9, \theta_2)$$ in polar coordinates. If $$\theta_1 - \theta_2 = \frac{\pi}{2},$$ then $$AB$$ find the distance

Apr 20, 2020

#1
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By polar coordinates, AB = sqrt(3^2 + 3*9 + 9^2) = 3*sqrt(13).

Apr 21, 2020
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Let $$A = (3, \theta_1)$$ and $$B = (9, \theta_2)$$ in polar coordinates. If  $$\theta_1 - \theta_2 = \dfrac{\pi}{2}$$, then  find the distance $$AB$$

Formula:
We have two points $$(r_1,\ \theta_1)$$ and $$( r_2,\ \theta_2)$$.
The distance between two polar coordinates $$\mathbf{d =\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1})}$$

$$\begin{array}{|rcll|} \hline d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1) } \\ d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos\Big(-(\theta_1 - \theta_2)\Big) } \\ d &=& \sqrt{r_1^2+r_2^2+2r_1r_2\cos(\theta_1 - \theta_2) } \\ d &=& \sqrt{3^2+9^2+2*3*9\cos\left( \dfrac{\pi}{2} \right) } \\ d &=& \sqrt{9+81+54\cos\left( \dfrac{\pi}{2} \right) } \quad | \quad \cos\left( \dfrac{\pi}{2} \right) = 0 \\ d &=& \sqrt{9+81+0 } \\ d &=& \sqrt{90} \\ d &=& \sqrt{9*10} \\ d &=& \sqrt{3^2*10} \\ \mathbf{d} &=& \mathbf{3\sqrt{10}} \\ \hline \end{array}$$

Apr 21, 2020
edited by heureka  Apr 21, 2020
edited by heureka  Apr 29, 2020
edited by heureka  Apr 29, 2020