What is the distance between $(4, \frac{7 π}{6})$ $(4 ,( 7 pi)/6 )$ and $(3, \frac{- π}{2})$ $(3 , ( - pi )/2 )$ ?

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What is the distance between $(4, \frac{7 π}{6})$ $(4 ,( 7 pi)/6 )$ and $(3, \frac{- π}{2})$ $(3 , ( - pi )/2 )$ ?

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Answer 1 · 2016-02-27T04:22:13

$\sqrt{}$ $sqrt$ 13

Explanation:

(x, y) = (r cos $θ$ $theta$ , r sin $θ$ $theta$ ) are components of the position vector ( r, $θ$ $theta$ ).
The radial vectors are ( $- 2$ $-2$ $\sqrt{}$ $sqrt$ 3, $- 2$ $-2$ ) and ( 0, $- 3$ $-3$ ). The vector between the given points is the difference ( $- 2$ $-2$ $\sqrt{}$ $sqrt$ 3, 1 ). .
The distance is the length of this vector = $\sqrt{}$ $sqrt$ 13.

Answer 2 · 2016-02-27T04:37:23

$3.606$ $3.606$

Explanation:

Point $(4, \frac{7 π}{6})$ $(4,(7pi)/6)$ in Cartesian coordinates represents $(4 \frac{cos (7 π)}{6}), 4 \frac{sin (7 π)}{6}))$ $(4cos(7pi)/6), 4sin(7pi)/6))$

i.e. $(4 \times (- \frac{\sqrt{3}}{2}), 4 \times (\frac{- 1}{2}))$ $(4xx(-sqrt3/2), 4xx((-1)/2))$ or $(- 2 \sqrt{3}, - 2)$ $(-2sqrt3, -2)$

Point $(3, - \frac{π}{2})$ $(3,-(pi)/2)$ in Cartesian coordinates represents $(3 cos (- \frac{π}{2}), 3 sin (- \frac{π}{2}))$ $(3cos(-pi/2),3sin(-pi/2))$

i.e. $(3 \times 0, 3 \times (- 1))$ $(3xx0, 3xx(-1))$ or $(0, - 3)$ $(0, -3)$

Hence distance between $(- 2 \sqrt{3}, - 2)$ $(-2sqrt3, -2)$ and $(0, - 3)$ $(0,-3)$

is $\sqrt{{(0 - (- 2 \sqrt{3}))}^{2} + {(- 3 - (- 2))}^{2}}$ $sqrt((0-(-2sqrt3))^2+(-3-(-2))^2$

= $\sqrt{{(2 \sqrt{3})}^{2} + {(- 1)}^{2}}$ $sqrt((2sqrt3)^2+(-1)^2$

= $\sqrt{12 + 1}$ $sqrt(12+1)$

= $\sqrt{13}$ $sqrt13$ = $3.606$ $3.606$

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What is the distance between $(4, \frac{7 π}{6})$ $(4 ,( 7 pi)/6 )$ and $(3, \frac{- π}{2})$ $(3 , ( - pi )/2 )$ ?

2 Answers

Explanation:

Explanation:

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