What are the components of the vector between the origin and the polar coordinate $(15, \frac{- 3 π}{4})$ $(15, (-3pi)/4)$ ?

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Answer 1 · 2017-11-29T18:21:24

$- \frac{15 \sqrt{2}}{2} ˆ i - \frac{15 \sqrt{2}}{2} ˆ j$ $-(15sqrt(2))/2hati-(15sqrt(2))/2hatj$

First convert the polar coordinate in a Cartesian coordinate. This can be done using:

$x = r cos (θ)$ $x=rcos(theta)$

$y = r sin (θ)$ $y=rsin(theta)$

$x = 15 cos (- \frac{3 π}{4}) = - \frac{15 \sqrt{2}}{2}$ $x= 15cos(-(3pi)/4)=-(15sqrt(2))/2$

$y = 15 sin (- \frac{3 π}{4}) = - \frac{15 \sqrt{2}}{2}$ $y= 15sin(-(3pi)/4)=-(15sqrt(2))/2$

Cartesian coordinates:

$(- \frac{15 \sqrt{2}}{2}, - \frac{15 \sqrt{2}}{2})$ $(-(15sqrt(2))/2 , -(15sqrt(2))/2 )$

Vector components.

$x = - \frac{15 \sqrt{2}}{2} ˆ i$ $x=-(15sqrt(2))/2hati$

$y = - \frac{15 \sqrt{2}}{2} ˆ j$ $y=-(15sqrt(2))/2hatj$

$:.$

$-(15sqrt(2))/2hati-(15sqrt(2))/2hatj$

What are the components of the vector between the origin and the polar coordinate (15, (-3pi)/4)(15,−3π4)(15, (-3pi)/4)?