Geometry with distance, vector.

Hello teacher~

Let c : I -> R^2 be a regular plane curve and
R a straight line in R^2.

If one can find a number t_0 in I such that
the distance from c(t) to R is greater than or equal to
the distance from c(t_0) to R, for all t in I,
and such that c(t_0) not in R,

then show that the tangent line of "c" at t_0 is parallel to R.

------------------------------------------------------------
hint)
If a in R and u in R^2 is a unit vector normal to the straight line R,
the function f : I -> R given by f(t) = < c(t) - a , u >
measures the (oriented) distance from the point c(t) to the line R.
---(***)

Since c(t_0) not in R, there is a neighbourhood of t_0 in I
in which f does not vanish.
Hence, f measures, up to a sign, the distance from the points of
(the trace of) c to R.
Then f has a minimum at t_0 in I and so f'(t_0) = 0.

-------------------------------------------------------------
I can't understand (***)part.
Namely,
What is <c(t) - a , u > ?
It's really the distance from the point c(t) to the line R ?


.

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