What are the components of the vector between the origin and the polar coordinate #(9, (3pi)/4)#?

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Answer 1 · 2017-08-08T01:45:00

#-(9sqrt2)/2hati + (9sqrt2)/2hatj#

What we'll do first is find the rectangular form of the given polar coordinate, using the equations

#ul(x = rcostheta#

#ul(y = rsintheta#

where in this case

So we have

#x = 9cos((3pi)/4) = -(9sqrt2)/2#

#y = 9sin((3pi)/4) = (9sqrt2)/2#

SInce the vector starts at the origin, these values are the #x#- and #y#-components of the vector, so we can write the vector as

#color(blue)(ulbar(|stackrel(" ")(" "< -(9sqrt2)/2, (9sqrt2)/2 >" ")|)#

Or

#color(blue)(ulbar(|stackrel(" ")(" "-(9sqrt2)/2 hati + (9sqrt2)/2 hatj" ")|)color(white)(aa)# (unit vector notation)