[proofplan]
View $\mathbb{P}^2_K$ as the projectivization of the [vector space](/page/Vector%20Space) $K^3$, so that projective lines are projectivizations of two-dimensional linear subspaces of $K^3$. If $L_1 = \mathbb{P}(A)$ and $L_2 = \mathbb{P}(B)$ with $A$ and $B$ two-dimensional subspaces, then distinctness of the projective lines means $A \neq B$. Since two distinct planes in a three-dimensional vector space have a one-dimensional intersection, their common projectivization consists of exactly one projective point.
[/proofplan]
[step:Represent the two projective lines by two-dimensional subspaces of $K^3$]
Let $E := K^3$. By the definition of the projective plane, $\mathbb{P}^2_K = \mathbb{P}(E)$, whose points are one-dimensional linear subspaces of $E$.
Since $L_1$ and $L_2$ are projective lines in $\mathbb{P}(E)$, there exist two-dimensional linear subspaces $A \subset E$ and $B \subset E$ such that
\begin{align*}
L_1 = \mathbb{P}(A)
\end{align*}
and
\begin{align*}
L_2 = \mathbb{P}(B).
\end{align*}
Here $\mathbb{P}(A)$ denotes the set of one-dimensional linear subspaces of $E$ contained in $A$, and similarly for $\mathbb{P}(B)$.
We claim that $A \neq B$. If $A = B$, then the one-dimensional subspaces contained in $A$ and in $B$ are the same, so $\mathbb{P}(A) = \mathbb{P}(B)$, contradicting $L_1 \neq L_2$. Thus $A$ and $B$ are distinct two-dimensional subspaces of $E$.
[/step]
[step:Show that the underlying two-dimensional subspaces meet in a one-dimensional subspace]
Because $A$ and $B$ are subspaces of the three-dimensional vector space $E$, their sum $A + B$ is a subspace of $E$. Since $A \neq B$ and both have dimension $2$, neither is contained in the other. Hence $A + B$ strictly contains $A$, so
\begin{align*}
\dim_K(A + B) > \dim_K A = 2.
\end{align*}
Since $A + B \subset E$ and $\dim_K E = 3$, it follows that
\begin{align*}
\dim_K(A + B) = 3.
\end{align*}
By the dimension formula for finite-dimensional subspaces,
\begin{align*}
\dim_K(A + B) = \dim_K A + \dim_K B - \dim_K(A \cap B).
\end{align*}
Substituting $\dim_K(A + B) = 3$ and $\dim_K A = \dim_K B = 2$ gives
\begin{align*}
3 = 2 + 2 - \dim_K(A \cap B).
\end{align*}
Therefore
\begin{align*}
\dim_K(A \cap B) = 1.
\end{align*}
[guided]
We need to translate the projective incidence question into a linear-algebra statement. The projective line $L_1$ consists of the one-dimensional subspaces of $E = K^3$ lying inside the two-dimensional subspace $A$, and $L_2$ consists of the one-dimensional subspaces lying inside the two-dimensional subspace $B$. Thus a projective point lies on both lines exactly when the corresponding one-dimensional vector subspace lies inside $A \cap B$.
The key linear-algebra fact is that two distinct two-dimensional subspaces of a three-dimensional vector space intersect in dimension $1$. We verify it directly. Since $A$ and $B$ are both two-dimensional and $A \neq B$, the subspace $B$ cannot be contained in $A$; if $B \subset A$, then the equality of dimensions would force $B = A$. Therefore $A + B$ strictly contains $A$, so
\begin{align*}
\dim_K(A + B) > 2.
\end{align*}
But $A + B$ is a subspace of $E = K^3$, so
\begin{align*}
\dim_K(A + B) \leq 3.
\end{align*}
Combining these two inequalities gives
\begin{align*}
\dim_K(A + B) = 3.
\end{align*}
Now apply the dimension formula for finite-dimensional subspaces:
\begin{align*}
\dim_K(A + B) = \dim_K A + \dim_K B - \dim_K(A \cap B).
\end{align*}
All spaces involved are finite-dimensional because they are subspaces of $K^3$. Substituting $\dim_K(A + B) = 3$ and $\dim_K A = \dim_K B = 2$ yields
\begin{align*}
3 = 2 + 2 - \dim_K(A \cap B).
\end{align*}
Solving this equality gives
\begin{align*}
\dim_K(A \cap B) = 1.
\end{align*}
Thus the common vector subspace underlying the intersection of the two projective lines is exactly one-dimensional.
[/guided]
[/step]
[step:Identify the unique projective point in the intersection]
Let $C := A \cap B$. From the previous step, $C$ is a one-dimensional linear subspace of $E$. Therefore $C$ represents a point of $\mathbb{P}(E) = \mathbb{P}^2_K$. Denote this projective point by
\begin{align*}
p := C \in \mathbb{P}^2_K.
\end{align*}
Since $C \subset A$, the point $p$ belongs to $\mathbb{P}(A) = L_1$. Since $C \subset B$, the point $p$ belongs to $\mathbb{P}(B) = L_2$. Hence
\begin{align*}
p \in L_1 \cap L_2.
\end{align*}
This proves existence.
[/step]
[step:Prove that no second projective point lies on both lines]
Let $q \in L_1 \cap L_2$ be any projective point. By the definition of $\mathbb{P}(E)$, the point $q$ is a one-dimensional linear subspace $Q \subset E$. Since $q \in L_1 = \mathbb{P}(A)$, we have $Q \subset A$. Since $q \in L_2 = \mathbb{P}(B)$, we have $Q \subset B$. Therefore
\begin{align*}
Q \subset A \cap B = C.
\end{align*}
Both $Q$ and $C$ are one-dimensional linear subspaces of $E$, so the inclusion $Q \subset C$ forces $Q = C$. Hence $q = p$.
Thus $L_1 \cap L_2$ contains exactly the single projective point $p$, completing the proof.
[/step]
