What are the components of the vector between the origin and the polar coordinate $(- 3, \frac{11 π}{6})$ $(-3, (11pi)/6)$ ?

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Answer 1 · 2018-07-22T18:15:56

$- \frac{3 \sqrt{3}}{2} ˆ i + \frac{3}{2} ˆ j$ $-\frac{3\sqrt3}{2}\hat i+3/2\hat j$

x & y components of vector $(r, θ) \equiv (- 3, \frac{11 π}{6})$ $(r, \theta)\equiv(-3, {11\pi}/6)$ are given as

$x = r cos θ$ $x=r\cos\theta$

$= - 3 cos (\frac{11 π}{6})$ $=-3\cos({11\pi}/6)$

$= - 3 cos (2 π - \frac{π}{6})$ $=-3\cos(2\pi-{\pi}/6)$

$= - 3 cos (\frac{π}{6})$ $=-3\cos({\pi}/6)$

$= - 3 \frac{\sqrt{3}}{2}$ $=-3\frac{\sqrt3}{2}$

$= - \frac{3 \sqrt{3}}{2}$ $=-\frac{3\sqrt3}{2}$

$y = r sin θ$ $y=r\sin\theta$

$= - 3 sin (\frac{11 π}{6})$ $=-3\sin({11\pi}/6)$

$= - 3 sin (2 π - \frac{π}{6})$ $=-3\sin(2\pi-{\pi}/6)$

$= 3 sin (\frac{π}{6})$ $=3\sin({\pi}/6)$

$= 3 \frac{1}{2}$ $=3\frac{1}{2}$

$= \frac{3}{2}$ $=\frac{3}{2}$

Hence the vector is

$= - \frac{3 \sqrt{3}}{2} ˆ i + \frac{3}{2} ˆ j$ $=-\frac{3\sqrt3}{2}\hat i+3/2\hat j$

What are the components of the vector between the origin and the polar coordinate (-3, (11pi)/6)(−3,11π6)(-3, (11pi)/6)?