Dimension of sum of 3 subspaces Let $W_1$, $W_2$ and $W_3$ be finite-dimensional subspaces of a vector space.

Show that it may happen that $W_i \cap W_j = 0$ for all $i \ne j$, but still
  $\dim(W_1 + W_2 + W_3) \ne \dim W_1 + \dim W_2 + \dim W_3 $.

I have a counterexample of three lines that represent each subspace that intersect at 0, but I don't really understand why that works. 
I know this property holds for two subspaces so I'm confused why it doesn't hold for three. 
 A: We know that $\dim(W_1 + W_2) = \dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$. Therefore, for three subspaces, $\dim(W_1+W_2+W_3) = \dim(W_1)+\dim(W_2+W_3)-\dim(W_1 \cap (W_2+W_3))$. Just because $W_1 \cap W_2 = 0$ and $W_1 \cap W_3 = 0$ does not mean that $W_1 \cap (W_2+W_3) = 0$.
A: I try to give an intuitive answer.
In the case with two subspaces, the condition $W_1 \cap W_2 = 0$ translates to the "linear independence of the two subspaces":
\begin{align*}
& \lambda_1 x_1 + \lambda_2 x_2 = 0 , \text{ where } x_1 \in W_1, x_2 \in W_2\\
\Longrightarrow & \lambda_1 x_1 = - \lambda_2 x_2  \in W_1 \cap W_2 = 0 \\
\Longrightarrow & \lambda_1 = \lambda_2 = 0 .
\end{align*}
While in  the case with three or more subspaces,  the pairwise "linearly independence" condition $W_i \cap W_j = 0$ for $i \neq j$ is not strong enough to ensure that each $W_i$ would "generate new/indepedent directions". We may need conditions like $(W_1 + \cdots + W_{j} ) \cap W_{j+1} =0$ for all $j \geq 1$ to ensure that $W_{j+1}$ is really independent from the subspace generated by the first $j$ subspaces. 
A: The condition $$\dim(W_1 + W_2 + W_3) = \dim W_1 + \dim W_2 + \dim W_3$$ is equivalent to the condition that the sum $W_1 + W_2 + W_3$ is a direct sum. We know that $W_1 + W_2 + W_3$ is a direct sum if and only if $W_1 \cap W_2 = 0$ and $(W_1 + W_2) \cap W_3 = 0$.
