What are the components of the vector between the origin and the polar coordinate #(-6, (17pi)/12)#?

1 Answer
Jun 20, 2016

The #x# component is #1.55#
The #y# component is #5.80#

Explanation:

The components of a vector are the amount the vector projects (i.e. points) in the #x# direction (this is the #x# component or horizontal component) and #y# direction (the #y# component or vertical component).

If the co-ordinates you'd been given were in Cartesian co-ordinates, rather than polar co-ordinates, you'd be able to read the components of the vector between the origin and the point specified straight from the co-ordinates, as they'd have the form #(x,y)#.

Therefore, simply convert into Cartesian co-ordinates and read off the #x# and #y# components. The equations that transform from polar to Cartesian co-ordinates are:
#x = r cos(\theta)# and
#y = r sin (\theta)#

The form of the polar co-ordinate notation you've been given is #(r, \theta) = (-6, \frac{17\pi}{12})#. So substitute #r = -6# and #\theta = \frac{17\pi}{12}# into the equations for #x# and #y#.

#x = -6 cos (\frac{17\pi}{12})#
#x = (-6) (-0.25882)#
#x = 1.5529#
#x \approx 1.55#

#y = -6 sin(\frac{17\pi}{12})#
#y = (-6)(-0.96593)#
#y = 5.7956#
#y \approx 5.80#

The co-ordinate of the point is therefore #(1.55,5.80)#.

The other end of the vector is at the origin, and so has co-ordinate #(0,0)#. The distance it covers in the #x# direction is therefore #1.55-0 = 1.55# and the distance it covers in the #y# direction is #5.80-0 = 5.80#.

The #x# component is #1.55# and the #y# component is #5.80#.

I highly recommend you have a look at this page on finding components of vectors. It works with polar and Cartesian co-ordinates, like you have done here, and has some diagrams that will make the process make sense. (There are lots of worked examples similar to this as well!)