anglevectors
Suppose we have 2 directional (not starting from the origin(0,0)) vectors u and v that are 2D vectors. Suppose (u and v) vectors can have any direction and can lie in any quadrant. The question is there a formula to find the angle between them?
Here is an image for more illustration:
After long search i found that i can use ATAN2() function for that, but the ATAN2() takes 2 positional vectors am i right?
Honestly, between the direction vector $ v=(10,5,-5)$ and the normal vector $( 2,-1,4)$ the angle is not acute, in fact it is obtuse (you can see that by a simple plot). 
However as you are asking about the angle between a line and a plane, so the you must take care of the orientation of the vectors you are working with.
In you case, to find the angle $ \theta $ you can do the following : when finding $\cos x$, apply $\arccos$ to find an angle $\phi$. Then subtract $\phi $ from $180$ to get $ \alpha=180 - \phi$ . Now $\theta = 90 - \alpha$.
The angle between two vectors is always less than 180 degrees.
If we are talking about angles greater than 180 degrees we need some other vocabulary. We need an orientation. Which vector is "on the right."
One approach:
Construct a matrix with the two vectors you are comparing in the first two columns, and the rest of the columns as principal component vectors.
The sign of the determinant will tell you the orientation of the two vectors.
Update
In 3-D
Suppose your vectors are $(x_1,x_2,x_3), (y_1,y_2,y_3)$
Then the the sign of $\det \begin{bmatrix} x_1&y_1&0\\x_2&y_2&0\\x_3&y_3&1\end{bmatrix}$ will give you the orientation between your two vectors from the point of view of the positive z axis.
If you wanted to extend the dimension.
$\det \begin{bmatrix} x_1&y_1&0&0\\x_2&y_2&0&0\\x_3&y_3&1&0\\x_4&y_4&0&1\end{bmatrix}$
Best Answer
Suppose that $u = (x, y)$ and $v = (a, b)$.
Let $$ w = (-y, x) $$ and compute \begin{align} c &= v \cdot u \\ s &= v \cdot w \end{align} Then the angle from $u$ to $v$ is exactly $atan2(s, c)$.
Details: The "dot product" of two vectors $(p, q)$ and $(r, s)$ is $pr + qs$. The length $\| x \|$ of a vector $(p, q)$ is $\sqrt{p^2 + q^2}$. So the formulas for $c$ and $s$ become \begin{align} c &= ax + by\\ s &= -ay + bx. \end{align}
The only problem that can arise is that $c = s = 0$, in which case the value returned by atan2 will not be meaningful. This only happens when either $u$ or $v$ is the zero vector (or both are!), in which case the angle between them is undefined anyhow.
One last thing: you've asked for the angle measured CLOCKWISE, but my answer gives the angle measured COUNTERclockwise, because that's a really well-established standard in mathematics. If you want a clockwise angle, you'll need to negate the result that my formula gives you.