Prove that every $3$ of $4$ vectors of $2$- planes are linearly independent.

Let $\: H_1=Span\{\vec{u_1},\: \vec{u_2}\}\:$ and $\:H_2=Span\{\vec{v_1},\: \vec{v_2}\}$ be subsets of $R^3$.

If $\: H_1$ and $\: H_2$ are two different planes, and  $H_1 ∩ H_2=l$ is a line $l=Span(\vec{e})$. Show that three of the vectors $\: \vec{u_1},\: \vec{u_2},\: \vec{v_1},\: \vec{v_2}$ are linearly independent.

We have: $\: H_1=Span\{\vec{u_1},\: \vec{u_2}\}\:$, $\:H_2=Span\{\vec{v_1},\: \vec{v_2}\}$, $\: H_1 ∩ H_2=l=Span(\vec{e})$, $\: H_1\neq H_2$

Since $\: H_1=Span\{\vec{u_1},\: \vec{u_2}\}\:$ is a plane, we know that $\vec{u_1}$ and $\: \vec{u_2}$ are linearly independent.

Since $\: H_2=Span\{\vec{v_1},\: \vec{v_2}\}\:$ is a plane, we know that $\vec{v_1}$ and $\: \vec{v_2}$ are linearly independent.

How do I continue?

Let's pretend I know that $\: \vec{v_1}∈H_1$. Then I could prove that $\{\vec{u_1},\: \vec{u_2},\: \vec{v_2}\}$ are linearly independent.

Since $\: H_1=Span\{\vec{u_1},\: \vec{u_2}\}\:$ is a plane, we know that $\vec{u_1}$ and $\: \vec{u_2}$ are linearly independent.

If $\{\vec{u_1},\: \vec{u_2},\: \vec{v_2}\}$ would be linearly dependent, then $\: \vec{v_2}$ would be in $H_1$ ($\: \vec{v_2}∈H_1$) and $H_1$ and $H_2$ would be the same plane ($H_1=H_2$), which they are not.

But what if I don't know that $\: \vec{v_1}∈H_1$?

Differentiate the two cases $v_1 \in H_1$ and $v_1 \notin H_1$
(a) If $v_1 \in H_1$ then you know how to proceed.
(b) If $v_1 \notin H_1$, then you are already finished, aren't you?