For location of a point, use
r ( cos theta, sin theta ) = ( x, y )r(cosθ,sinθ)=(x,y).
Let any vec v = vec r→v=→r
= r < cos theta, sin theta >=r<cosθ,sinθ>
<< x-component x, y-component y >>
= < x, y >=<x,y>.
vec r→r from the ( origin ) pole, r = 0 rArr x = y =0,
is the like-parallel and equal to vec v→v.
Length r = length v.
thetaθ given by vec v→v.
Now, the unit vector
in the direction vec r→r ( for that matter vec v→v ) is
vec u = vec r/r = 1/sqrt ( x^2 +y^2) < x, y >→u=→rr=1√x2+y2<x,y>
In other words,
any direction can be represented by
unit vector vec u = vec r/r→u=→rr, in the direction of vec v→v.
Note that length of the vector vec r, r = sqrt ( x^2 + y^2 ) >= 0→r,r=√x2+y2≥0.
If r = 1, vec r = vec u→r=→u.
Unit vector vec i = 1 < 1, 0 >→i=1<1,0> for #theta = 2kpi, gives x-positive x-axis.
Unit vector - vec i−→i = 1 < -1, 0 >for #theta = ( 2kp + 1 )pi,
gives x-negative x-axis.
Unit vector vec j = 1 < 0, 1 >→j=1<0,1> for #theta = ( 2k +1/2 )pi, gives y-positive y-axis.
Unit vector - vecj = 1 < 0, - 1 >−→j=1<0,−1> for #theta = ( 2k + 3/2 )pi, gives y-negative y-axis.
It is evident that vec v = < x, y >→v=<x,y>
is a short form for the sum of the component vectors
x veci + y vecj = r ( cos theta veci + sin theta vecj)x→i+y→j=r(cosθ→i+sinθ→j)
Graph of the like-parallel and equal vector vecr→r at O ( 0, 0 )O(0,0),
for vec v = < 3, -4 >→v=<3,−4>, positioned elsewhere, in the direction
#theta = arctan( -4/3 ) .
graph{3y+4x=0[0 8 -4 0]}
The direction is away from O.
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