A question about analytic geometry (vectors) Let $a , b, c$ be three vectors such that $a+3b+c = o$ where $o$ is $(0,0,0)$ vector. we also know $|a| = 3 , |b| = 4 , |c| =6$. we want to find $a.b + a.c + b.c$
I know we can solve it by saying $a+b+c = -2b$ and then squaring the sides. and at last we get the answer 3/2 . But my problem is that is it even possible for $|a| = 3 , |b| = 4 , |c| =6$ to be such that $a+3b+c = o$ ? I think it is not possible because of triangle inequality. If it is possible, how? and If it is not, why do we get an answer for it? 

Dot the equation with each of the vectors
\begin{eqnarray*}
\mid a \mid^2 +3 a \cdot b +a \cdot c =0 \\
a \cdot b +3 \mid b \mid^2 +b \cdot c =0 \\
 a \cdot c +3 b \cdot c+\mid c \mid^2  =0 \\
\end{eqnarray*}
Now add the first and third equations and subtract the second $2(a \cdot b +a \cdot c +b \cdot c)+9+36-48=0$ ...
