vectors
I have two vectors:
$A = i-3j+2k$ and
$B = -3i+4j-k.$
I have already found the angle between the two to be ~$157.6$ degrees, but I am required to find the smallest angle between the two – I am not sure what that means exactly and can't seem to find the answer online. I would be glad if you could help me out with that.
The dot product of two normalized vectors is equal to the cosine of the angle between them. In general $$\cos \phi = \frac{a\cdot b}{|a||b|},$$ since your vectors are normalized, $|a|=|b|=1$ and $\phi = \arccos(a\cdot b)$
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 $\theta = 90 - \alpha$.
Best Answer
They are position vectors in $\mathbb{R}^3$. Two vectors describe a plane so I presume you are being asked to find the angle between the two vectors in the plane they describe.
I've added a small illustration (done in Geogebra) to show what I mean. The two points $A$ and $B$ are described by the vectors you gave. The blue plane is described by $A, B$ and the origin and the angle between these two vectors is $153^{\circ}$ (not sure if the angle is properly visible). You can compute this angle via the dot product. If you used the dot product and got to $157.6^{\circ}$ there may be a rounding error of some description in your calculation. Either way, I believe this may be what your teacher is looking for. Does this help?