7. Given $\boldsymbol{a}$ and $\boldsymbol{b}$ are two mutually perpendicular unit vectors, and $c \cdot a=c \cdot b=1$. Then for any positive real number $t$, $\left|c+t a+\frac{1}{t} b\right|$ has the minimum value of $\qquad$.

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$7.2 \sqrt{2}$.
Let $c=\lambda a+\mu b$. Then
$$
\begin{array}{l}
\boldsymbol{c} \cdot \boldsymbol{a}=\lambda|\boldsymbol{a}|^{2}+\mu \boldsymbol{a} \cdot \boldsymbol{b}=\lambda=1, \\
\boldsymbol{c} \cdot \boldsymbol{b}=\lambda \boldsymbol{a} \cdot \boldsymbol{b}+\mu|\boldsymbol{b}|^{2}=\mu=1 .
\end{array}
$$

Thus, $c=a+b$.
So $c^{2}=(a+b)^{2}=a^{2}+b^{2}+2 a \cdot b=2$.
$$
\begin{array}{l}
\text { Also }\left|c+t a+\frac{1}{t} b\right|^{2}=\left(c+t a+\frac{1}{t} b\right)^{2} \\
=c^{2}+t^{2} a^{2}+\frac{1}{t^{2}} b^{2}+2 t c \cdot a+\frac{2}{t} c \cdot b+2 t a \cdot \frac{1}{t} b \\
=2+t^{2}+\frac{1}{t^{2}}+2 t+\frac{2}{t} \\
=\left(t+\frac{1}{t}+1\right)^{2}-1 \\
\geqslant(2+1)^{2}-1=8(t>0),
\end{array}
$$

Therefore, $\left|c+t a+\frac{1}{t} b\right| \geqslant 2 \sqrt{2}$, when $t=1$, the equality holds in the above expressions.
Hence, the minimum value sought is $2 \sqrt{2}$.