I have been independently working proofs out of a vector calculus book I own and would like feedback on one of them; if it's right; if it is right, could it be improved; etc. Here is the problem: (Also, I apologize in advance, I don't have software to let me type math equations. Please read numbers and lower-case letters attached to the right of capital letters as subscripts.)
Let $A_1, A_2,\ldots,A_r$ be non-zero vectors which are mutually perpendicular, or in other words, $A_i\cdot A_j = 0$ iff $i\ne j$. Let $c_1,c_2,\ldots,c_r$ be real numbers such that $c_1A_1 + \ldots + c_rA_r = \mathbf{0}$. Prove that all $c_i = 0$.
My attempt: $c_1A_1 + \ldots + c_rA_r = \mathbf{0}$
(take the dot product of each side and $A_r$)
$A_r\cdot(c_1A_1 + \ldots + c_rA_r) = A_r \cdot\mathbf{0}$
reduces to
$c_r(A_r\cdot A_r) = 0$
(divide both sides by $A_r\cdot A_r$)
$c_r = 0.$
Since $r$ is arbitrary, any $c_r$ must be zero. QED.
Is this a sufficient proof of the question asked? Thanks in advance.