7. Let the side length of the equilateral triangle $\triangle A B C$ be $1, t$ be any real number. Then the minimum value of $|\overrightarrow{A B}+t \overrightarrow{A C}|$ is $\qquad$
7. $\frac{\sqrt{3}}{2}$.

Let $a=\overrightarrow{A B}, b=\overrightarrow{A C}$. Then $|a|=|b|=1$, and the angle between $a$ and $b$ is $60^{\circ}$. Therefore,
$$
\begin{array}{l}
|\overrightarrow{A B}+t \overrightarrow{A C}|^{2}=|a+t b|^{2} \\
=a^{2}+t^{2} b^{2}+2 t a \cdot b=1+t^{2}+2 t \cos 60^{\circ} \\
=1+t^{2}+t=\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4} .
\end{array}
$$

Thus, $|\overrightarrow{A B}+t \overrightarrow{A C}| \geqslant \frac{\sqrt{3}}{2}$.

The equality holds if and only if $t=-\frac{1}{2}$.