vectors
Two planes have equations $x + 3y − 2z = 4$ and $2x + y + 3z = 5$. The planes intersect in the straight line l.
$(i)$ Calculate the acute angle between the two planes.
How do you this, I know from the normal form of a vector line is the dot product, what about in cartesian form?
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 $\theta = 90 - \alpha$.
Let: $ f'(x) =(x^3)' = 3x^2$ $ g'(x) =(x^2)' = 2x$
Next solve equation:
$x^2 = x^3$
$x^2-x^3 = 0$
$x^2(1-x) = 0$
Hence intesection points are at $x=0$ and $x=1$
At $x=0$: $f'(0) = 0$ and $g'(0) $ - derivatives are equal hence angle is 0.
At $x=1$ : $f'(1)= 3$ and $g'(1)=2$ hence the angle is $\arctan(3)-\arctan(2) \approx 0.141897$ because the slope of the line equals $\tan(\alpha)$ where $\alpha$ is angle of inclination of line. (https://en.wikipedia.org/wiki/Linear_function)
Best Answer
Given the equation of a plane $Ax + By + Cz = d$, the normal vector of the plane is (A,B,C), i.e the vector perpendicular to that plane. Using this you can obtain the two normal vectors, and then apply their escalar product:
$$A_1*A_2 + B_1*B_2+C_1*C_2=|n_1|*|n_2|*cos\alpha$$
where $n_1,n_2$ denote the normal vectors and |·| is the module of the vector.