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?
 A: If some $\alpha_i\neq 0$, say $\alpha_1$ without loss of generality, then $u=-\frac{\alpha_2}{\alpha_1}v-\frac{\alpha_3}{\alpha_1}w$. This means that $u$ lies in some plane containing the origin, $v$ and $w$.
A: It is all a question about whether the vectors $u$, $w$, and $w$ are linearly independent. Lets assume that they are linearly dependent. Then with out loss of generality you can assume that $u = a_1v+ a_2w$ for $(a_2,a_2) \neq (0,0)$. That means that the span of $u,v,w$ is the span of $v$ and $w$. Assuming that these are linearly dependent, then the span of $u,v,w$ is a plane:
$$
su + tw,\quad s,t\in \mathbb{R}.
$$
This obviously contains the origin. 
If $v$ and $w$ are also linearly dependent, then the span of $u,v,w$ is a line that passes through the origin, so any plane containing the three points will contain the origin.
We have proved that if the vectors are dependent, then there is a plane that contains all of them (the points) that also contains the origin. Reversely, if the three vectors satisfy that any plane containing all points do not contain the origin, the three vectors must be linearly independent.
A: 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 !
