Properties of the dot product

From $$\vec{A}\cdot\vec{C}=\vec{B}\cdot\vec{C}$$ you can conclude $$\vec{A}\cdot\vec{C}-\vec{B}\cdot\vec{C}=0$$ or $$(\vec{A}-\vec{B})\cdot\vec{C}=0.$$ However, this does not necessarily mean $\vec{A}-\vec{B}=\vec{0}$.

You can only conclude (from the definition of the dot product) that $\vec{A}-\vec{B}$ is perpendicular to $\vec{C}$.

If for given $\vec A$ and $\vec B$ the equality $\vec A\cdot\vec C = \vec B\cdot\vec C$ holds for all vectors $\vec C$, or at least for a set of generators (say, a basis), then we can conclude that the two vectors are equal, otherwise we can't.

I will try to make it plausible: If we take the standard basis $\{\vec e_x, \vec e_y, \vec e_z\}$ for vector $\vec C$, then we get

$$\vec A\cdot\vec e_x = \vec B\cdot\vec e_x$$ $$\vec A\cdot\vec e_y = \vec B\cdot\vec e_y$$ $$\vec A\cdot\vec e_z = \vec B\cdot\vec e_z$$

But $\vec A \cdot \vec e_i=A_i$, the i-th component of the vector. So we have just shown that $A_x = B_x$, $A_y=B_y$ and $A_z=B_z$ and thus $\vec A = \vec B$.

On the other hand, let us assume that the equality holds for the first two basis vectors but doesn't for the last one, $\vec e_z$, then we know that the first two components of $\vec A$ and $\vec B$ coincide but the z components don't and thus the two vectors aren't equal.

The approach works for any basis, not necessarily the standard basis. Then you will get components with respect to the given basis -- if they all coincide, then the vectors also coincide.

It's not a rigorous proof but maybe helps to make the statement intuitive.

Here is a proof. If $\mathbf{A}\cdot\mathbf{C} =\mathbf{B}\cdot\mathbf{C}$ for all $\mathbf{C}$, then $(\mathbf{A}-\mathbf{B})\cdot\mathbf{C} = 0$ for all $\mathbf{C}$. In particular we can choose $\mathbf{C} = \mathbf{A}-\mathbf{B}$ so that $(\mathbf{A}-\mathbf{B})\cdot(\mathbf{A}-\mathbf{B})=0$. Since the dot product is positive definite, $\mathbf{v} \cdot \mathbf{v} = 0$ only if $\mathbf{v} = 0$. We conclude $\mathbf{A} - \mathbf{B} = 0$, so $\mathbf{A} = \mathbf{B}$.