accelerationdifferentiationkinematicsmathematicsvectors
My physics teacher told us that we can’t divide vectors, that vector division has no physical meaning or significance. How about this: $$a = vdv/dx.$$
It says acceleration vector equals velocity (as a function of $x$) times $dv$ ‘divided’ by $dx$.
Here both $dv$ and $dx$ are vectors. How do I make sense of it? Because vector division doesn’t exist in physics right?
I get the physical significance of vector addition & subtraction. But I don't understand what do dot & cross products mean?
Perhaps you would find the geometric interpretations of the dot and cross products more intuitive:
The dot product of A and B is the length of the projection of A onto B multiplied by the length of B (or the other way around--it's commutative).
The magnitude of the cross product is the area of the parallelogram with two sides A and B. The orientation of the cross product is orthogonal to the plane containing this parallelogram.
Why can't vectors be divided?
How would you define the inverse of a vector such that $\mathbf{v} \times \mathbf{v}^{-1} = \mathbf{1}$? What would be the "identity vector" $\mathbf{1}$?
In fact, the answer is sometimes you can. In particular, in two dimensions, you can make a correspondence between vectors and complex numbers, where the real and imaginary parts of the complex number give the (x,y) coordinates of the vector. Division is well-defined for the complex numbers.
The cross-product only exists in 3D.
Division is defined in some higher-dimensional spaces too (such as the quaternions), but only if you give up commutativity and/or associativity.
Here's an illustration of the geometric meanings of dot and cross product, from the wikipedia article for dot product and wikipedia article for cross product:
There's two possible views here.
The second view is to understand the difference between displacement and position, and that position in general is not a vector to begin with. Position is a point, or better, a numerical (i.e. coordinates) way of denoting points, and while you can "get away with" treating it as a vector in some cases, you cannot in others, e.g. on a curved manifold. Displacements are vectors derived from positions.
Positions and vectors, are, if you like (and there is actually a foundational mathematical theory, but unfortunately it doesn't seem to be too well used, called "type theory" for this) belong to different "data types", with different semantic denotations and different operations you can and cannot perform them. In paritcular, while both positions and vectors can be represented as tuples of real numbers drawn out of the sets $\mathbb{R}^n$, you cannot add or scale positions, but you can vectors.
Indeed, typically you can't "do" anything to positions except to compare them in some fashion - whether that be in terms of linear ordering, as in one dimension, i.e. the real line - or in terms of measuring the distance between them (which is what a metric space or metrical manifold [Riemannian etc.] is about). But you can "do" things to vectors - add, scale, etc. .
And this, I believe, is the better way to make sense of what is going on here. In the particular case of positions in Euclidean space, which is what is called an "affine space", you do also have the option of subtracting, to get vectorial displacements $\mathbf{d} = Q - P$ for points $P$ and $Q$. However, this does not work on curved manifolds in any natural way, because of their nonlinear geometry - unless you are talking about a displacement who is so small that the curvature can be neglected and the manifold considered flat. That is, on curved manifolds, you can only differentiate a curvilinear displacement, which is in most proper sense a curve, or function, between the points, i.e. $\gamma: [0, 1] \rightarrow M$, such that $\gamma(0) = P$ and $\gamma(1) = Q$.
And angular position, turns out, most honestly "lives" in a curved manifold, and does not live in Euclidean space - instead, it "lives" on a manifold that is a form of the "real projective space" $\mathbf{RP}^3$, and can be constructed more explicitly using a convention such as the Euler angles or Tait-Bryan angles. That is, a point in angular position manifold is given by $$(\psi, \theta, \phi)$$ where these are the three Euler angles (personally, I find the Tait-Bryan angles ['roll/pitch/yaw'] more intuitive).
That said, there is, then, also a corresponding rigorous notion of angular displacement - but this is because the way these angular positions act to rotate the object is through the action of the corresponding rotation matrix, and those rotation matrices can be composed. The resulting structure is thus an algebraic group, not a vector space; it is the Lie group $\mathrm{SO}(3)$. And you can write its displacement elements in the above form, too, with angles, but the composition in terms of the angles is not simple addition thereof ala a vector space, hence they are not "vectors", but their own thing which is specific to angle space. Nonetheless, they can still be differentiated to vectors, as you have observed - this is because any curve $\gamma$ on a differentaible manifold will differentiate to a vector in the tangent spaces $T_xM$.
Best Answer
The statement $a = v (dv/dx)$ only holds in that form for one-dimensional motion, where the quantities $v$ and $x$ are just numbers rather than vectors. It follows from the chain rule, if we view $v$ as a function of $x$ instead of as a function of $t$: $$ a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v. $$
If you're doing 2-D or 3-D motion, you can still do something similar, but you have to let $\vec{v}$ be a function of $x$, $y$, and $z$, since $\vec{v}$ can change as each of these quantities changes. This means that you need to use multi-variable calculus to write out an equivalent statement. For example, we have $$ a_x = \frac{dv_x}{dt} = \frac{\partial v_x}{\partial x} \frac{dx}{dt} + \frac{\partial v_x}{\partial y} \frac{dy}{dt} + \frac{\partial v_x}{\partial z} \frac{dz}{dt} \\= \frac{\partial v_x}{\partial x} v_x + \frac{\partial v_x}{\partial y} v_y + \frac{\partial v_x}{\partial z} v_z. $$ As you can see, we're never "dividing by" the entire vector $\vec{x}$ when we take these derivatives; we only ever "divide by" its components $x$, $y$, or $z$, which is a valid mathematical operation.