Prove that the acceleration is either zero or parallel to the velocity given the curve lies on a plane I'm stuck on this problem (section 14.9 exercise 13 of Apostol Calculus) :

A particle moves along a curve in such a way that the velocity vector makes a constant angle with a
given unit vector $\mathbf{c}$ . If the curve lies in a plane containing $\mathbf{c}$, prove that the acceleration vector  is either zero or parallel to the velocity

My attempt :
Define a unit vector $\displaystyle  \mathbf{T} = \frac{\mathbf{v}}{\|\mathbf{v}\|}$,  so $\displaystyle \mathbf{T} \ ' =  \frac{\mathbf{a}\|\mathbf{v}\|-\mathbf{v}\|\mathbf{v}\|'}{\|\mathbf{v}\|^2}$.
Then, because of the angle between the velocity $v$ and $c$ is constant we have
$$\mathbf{T} \cdot \mathbf{c} = \cos \alpha \implies \mathbf{T}\ ' \cdot \mathbf{c} = 0 \implies \mathbf{T} \ '=\mathbf{0} \  \lor \  \mathbf{T}\ ' \bot \ \mathbf{c}  $$
Also,
$$\mathbf{T} \cdot \mathbf{T}\ = 1 \implies \mathbf{T} \ ' \cdot v = 0 \implies    \mathbf{T} \ '=\mathbf{0} \  \lor \  \mathbf{T}\ ' \bot \ \mathbf{v}$$
If it is true that $ \mathbf{T}\ '=\mathbf{0} $ then $\displaystyle \mathbf{a}=\frac{\|\mathbf{v}\|'}{\|\mathbf{v}\|}\mathbf{v}  $ ,  so we have done.
Edit:
But as @user pointed out, $\mathbf{T} '$ cannot be perpendicular to both $\mathbf{v}$ and $\mathbf{c}$ because $\mathbf{v}$ and $\mathbf{a}$ have to lie on the same plane as $\mathbf{c}$ so does $\mathbf{T} '$, but this means it must be $\mathbf{0}$ because there isn't a non zero vector on a plane perpendicular to $\mathbf{v}$ and $\mathbf{c}$.
I think all of my problem was that I was not aware of the definition of vector lying on a plane, i thought just the spike of it was included in the plane.
 A: Consider another hypothetical particle, which starts at the origin and its position vector (which is therefore also the displacement vector) is same as the velocity vector of the given particle, at any given time.
Now since the position vector of this particle is always at constant angle to $\mathbf c$, so its trajectory is a cone, with origin as apex and $\mathbf c$ to be the axis.
Also, since the direction of movement of any particle is actually the direction of its velocity, so the velocity of the given particle (and hence the displacement of this particle) is in a plane containing $\mathbf c$.
Hence by the geometry of a cone, this particle moves in a straight line (since changing direction would either make its trajectory non-coplanar with $\mathbf c$, or will change the angle with $\mathbf c$). Hence the velocity of this particle (and thus the acceleration of the given particle) is always in the same direction as the displacement of this particle (velocity of given particle), or zero. This is what we wanted to show.
A: Since the curve and unit vector are in the same plane, let that plane be xy plane.Set original velocity  $\hat v =<x,y,0 >$, unit vector = $\hat u$=<0,1,0>,  $\theta$ is the angle between $\hat v$ and $\hat u$ . After the acceleration, velocity vector becomes  $\hat v$=<x',y',z'>.
Since the angle between  $\hat v'$ and $\hat u$ need to be $\theta$, $\hat v $ and $\hat v'$ are parallel to each other, they are scaled by a scalar value m,   or
$\hat v' = m\hat v$ =$<mx,my,0>$,
the velocity change is
$\Delta \hat v$=$<mx-m,my-y,0>$ =(m-1)$\hat v$.
Acceleration $\hat a$ =  $\frac{\Delta v}{\Delta t}$=$\frac{(m-1) \hat v}{\Delta t}$,
$\hat a$  is either parallel to  $\hat v$ (m$\neq 1$)or $\hat 0$ ($ m=1$).
