Show that $\alpha_1u+\alpha_2v+\alpha_3w=0\Rightarrow\alpha_1=\alpha_2=\alpha_3=0$ Let $u, v, w$ be three points in $\mathbb R^3$ not lying in any plane containing the origin. Would you help me to prove or disprove: $\alpha_1u+\alpha_2v+\alpha_3w=0\Rightarrow\alpha_1=\alpha_2=\alpha_3=0.$ I think this is wrong since otherwise Rank of the Coefficient Matrix have to be 3. But for$u_1=(1,1,0),u_2=(1,2,0),u_3=(1,3,0)$, (Rank of the corresponding Coefficient Matrix)$\neq 3$. 
Am I right?

With $u_1=(1,1,0)$, $u_2=(1,2,0)$, $u_3=(1,3,0)$
Let $A = \begin{pmatrix}
  u_1 \\
  u_2 \\
  u_3
 \end{pmatrix} = \begin{pmatrix}1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 3 & 0 \end{pmatrix}$ 
we have $\det A = 0$ $\implies$ $u_1, u_2, u_3$ is linearly dependent  $\implies$ you're wrong !
