The scalar multiple of one vector with two other vectors Prove or disprove:
$(\vec{A}\cdot\vec{B}=\vec{A}\cdot\vec{C}) \rightarrow \vec{B}=\vec{C}$
When all vectors are not $\vec{0}$ and are in $\mathbb{R}^n$
 A: Let the inner product $\langle \,\,, \, \rangle$ be the standard dot product in $\mathbb{R}^n$ (i.e. for $A,B \in \mathbb{R}^n$, $A \cdot B = \langle A,B\rangle$). Now, recall that the dot product is non-degenerate. This means that for a fixed $A \in \mathbb{R}^n$, if $\langle A, B \rangle = 0$ for all $B$, then $A=0$. (Here $0$ denotes the vector in $\mathbb{R}^n$ with every component $0$). So, to prove your statement, we rewrite the left-hand side of the implication using inner product notation: 
\begin{equation}
\langle A, B\rangle = \langle A, C \rangle \Rightarrow \langle A, B-C \rangle=0
\end{equation}
Now, if $\langle A, B-C \rangle = 0$, then all this says is that $B-C$ is orthogonal to $A$, and you can choose a non-zero vector in $\mathbb{R}^n$ such that this is true. However, if we consider that $\langle A, B-C \rangle = 0$ for all $A \in \mathbb{R}^n$, then $B-C$ must be zero by the non-degenerate property of $\langle \, , \rangle$.
Added in response to OP's comment:
If you don't consider inner products, you can imagine the equation as $$A \cdot B = A \cdot C \Rightarrow A \cdot (B-C)=0$$Assume $B \ne C$. If the above equation holds for all $A \in \mathbb{R}^n$, then let $A=B-C$. Then we have $(B-C)\cdot (B-C)=0$ or $||B-C||^2=0$. This implies that $||B-C||=0$ which shows that $B-C$ must be the zero vector (this is the only vector in $\mathbb{R}^n$ with norm (length) of $0$). However, since this implies that $B=C$ this is a contradiction, since we assumed $B \ne C$. Therefore, our assumption that $B \ne C$ was wrong, and $B=C$.
A: If $A$ is an arbitrary vector then you can consider it to be different basis vectors $\left\{ {{b_1},{b_2},...,{b_n}} \right\}$ for $\mathbb{R}^n$,  then your assumption implies that
$$\left( {B - C} \right) \cdot {b_i} = 0\,\,\,\,\,\,\,\,\,\,\,\,,i = 1,2,...,n$$
and hence $B-C$ is a zero vector as its dot products with basis vectors are zero and hence $B=C$.
