Let $U$ be any finite set of points in the plane, and $U$ contains $k(k \geqslant 3)$ non-collinear points. Find the minimum value of $|U-U|$.
First, we provide the construction:
When $k=2 l+1\left(l \in \mathbf{Z}_{+}\right)$, let
$$
\begin{aligned}
U= & \{(i, 0) \mid 1 \leqslant i \leqslant l, i \in \mathbf{Z}\} \cup \\
& \{(i, 1) \mid 1 \leqslant i \leqslant l+1, i \in \mathbf{Z}\} .
\end{aligned}
$$

When $k=2 l(l \in \mathbf{Z}, l \geqslant 2)$, let
$$
U=\{(i, j) \mid 1 \leqslant i \leqslant l, 1 \leqslant j \leqslant 2, i 、 j \in \mathbf{Z}\} \text {. }
$$

Notice that, if $v \in U-U$, then $-v \in U-U$.
Thus, $2|I(U-U) \backslash\{0\}|$
$\Rightarrow|U-U| \equiv 1(\bmod 2)$.
Therefore, it suffices to prove: $|U-U| \geqslant 3 k-3$.
We proceed by induction on $k$.
When $k=3$, the three points in the set $U$ form a triangle, hence,
$$
|U-U|=7 \geqslant 3 k-3 \text {. }
$$

Assume $k \geqslant 4$ and the conclusion holds for $3,4, \cdots, k-1$.
Now consider the case for $k$.
Since the points in the set $U$ are not all collinear, the convex hull of $U$ is a convex polygon, which we still denote as $U$. Take a line $L$ through an edge of $U$, and take a point in $U$ farthest from the line $L$ to draw a line $L^{\prime}$ parallel to $L$. Thus, the remaining points in $U$ are on the lines $L$ and $L^{\prime}$ or between them.

Let the number of intersection points of $U$ with lines $L$ and $L^{\prime}$ be $t$ and $s$, respectively, denoted as
$|L \cap U|=t,\left|L^{\prime} \cap U\right|=s$.
Without loss of generality, assume $t \geqslant s$, then $t \geqslant 2, s \geqslant 1$.
Let the points in $L \cap U$ from left to right be $A_{1}, A_{2}$, $\cdots, A_{t}$, and the points in $L^{\prime} \cap U$ from left to right be $B_{1}, B_{2}$, $\cdots, B_{s}$.
Consider the following two sets of vectors
(1) $\overrightarrow{A_{1} B_{1}}, \overrightarrow{A_{1} B_{2}}, \cdots, \overrightarrow{A_{1} B_{s}}, \overrightarrow{A_{2} B_{1}}, \cdots, \overrightarrow{A_{t} B_{1}}$;
(2) $\overrightarrow{B_{1} A_{1}}, \overrightarrow{B_{2} A_{1}}, \cdots, \overrightarrow{B_{s} A_{1}}, \overrightarrow{B_{1} A_{2}}, \cdots, \overrightarrow{B_{1} A_{i}}$.
It is easy to see that the following two conclusions hold:
(i) The projections of the vectors in (1) and (2) on $L$ are distinct;
(ii) The projections of the vectors in (1) and (2) in the direction perpendicular to $L$ are also distinct.

Thus, (1) and (2) together have $2(t+s-1)$ distinct vectors.
Hence, $\left| \pm\left(L \cap U-L^{\prime} \cap U\right)\right| \geqslant 2(t+s-1)$.
Note that, if $A$ and $B$ are two non-empty sets, then
$$
|A+B| \geqslant|A|+|B|-1 \text {. }
$$

When $k=t+s$, i.e., there are no other points between lines $L$ and $L^{\prime}$,
$$
\begin{array}{l}
|U-U| \geqslant|L-L|+\left| \pm\left(L \cap U-L^{\prime} \cap U\right)\right| \\
\geqslant 2 t-1+2 t+2 s-2=4 t+2 s-3 \\
\geqslant 3 t+3 s-3=3 k-3 .
\end{array}
$$

When $k>t+s$, i.e., there are other points between lines $L$ and $L^{\prime}$, let $U^{\prime}=U \backslash L^{\prime}$. Since line $L$ contains at least two points of $U^{\prime}$, and there is a point outside line $L$, the points in $U^{\prime}$ are not all collinear.
By the induction hypothesis,
$$
\left|U^{\prime}-U^{\prime}\right| \geqslant 3(k-s)-3 \text {. }
$$

Also, the projection length of $\pm\left(L \cap U-L^{\prime} \cap U\right)$ in the direction perpendicular to line $L$ is greater than the projection length of $U^{\prime}-U^{\prime}$ in the direction perpendicular to line $L$, then
$$
\begin{array}{l}
|U-U| \geqslant\left|U^{\prime}-U^{\prime}\right|+\left| \pm\left(L \cap U-L^{\prime} \cap U\right)\right| \\
\geqslant 3(k-s)-3+2(t+s-1) \\
=3 k-3+(t-s)+(t-2) \\
\geqslant 3 k-3 .
\end{array}
$$

In summary, $|U-U| \geqslant 3 k-3$.
Thus, $|U-U|_{\min }=\left\{\begin{array}{l}3 k-2, k \text { is odd; } \\ 3 k-3, k \text { is even. }\end{array}\right.$