7. In space, four non-collinear vectors $\overrightarrow{O A} 、 \overrightarrow{O B}$ 、 $\overrightarrow{O C} 、 \overrightarrow{O D}$ have pairwise angles of $\alpha$. Then the size of $\alpha$ is $\qquad$
7. $\arccos \left(-\frac{1}{3}\right)$.

Let's assume $|\overrightarrow{O A}|=|\overrightarrow{O B}|=|\overrightarrow{O C}|=|\overrightarrow{O D}|=1$ and
$$
\overrightarrow{O D}=a \overrightarrow{O A}+b \overrightarrow{O B}+c \overrightarrow{O C} \text {. }
$$

Taking the dot product of both sides with $\overrightarrow{O D}$, we get
$$
\begin{array}{l}
1=a \cos \alpha+b \cos \alpha+c \cos \alpha \\
=(a+b+c) \cos \alpha \\
\Rightarrow a+b+c=\frac{1}{\cos \alpha} .
\end{array}
$$

If we take the dot product of both sides of equation (1) with $\overrightarrow{O A}$, we get
$$
\begin{array}{l}
\cos \alpha=a+b \cos \alpha+c \cos \alpha \\
\Rightarrow a=(1-b-c) \cos \alpha .
\end{array}
$$

Similarly, $b=(1-c-a) \cos \alpha$,
$$
c=(1-a-b) \cos \alpha \text {. }
$$

Adding the three equations and using equation (1), we get
$$
\begin{array}{l}
\frac{1}{\cos \alpha}=\left(3-\frac{2}{\cos \alpha}\right) \cos \alpha \\
\Rightarrow \cos \alpha=-\frac{1}{3}(\cos \alpha=1 \text { is discarded) } \\
\Rightarrow \alpha=\arccos \left(-\frac{1}{3}\right) .
\end{array}
$$