7. Given planar vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ satisfying $|\boldsymbol{a}|=|\boldsymbol{b}|=1,|\boldsymbol{c}|=5$, and $\boldsymbol{a} \cdot \boldsymbol{c}=3, \boldsymbol{b} \cdot \boldsymbol{c}=4$, then for any real number $t, z=|c-t a-b|$ the minimum value is $\qquad$.
7. 3 Detailed Explanation: $\because|\boldsymbol{a}|=|\boldsymbol{b}|=1, \therefore \cos \angle C O A=\frac{3}{5}, \cos \angle C O B=\frac{4}{5}$; $\therefore \sin \angle C O A=\frac{4}{5}, \sin \angle C O B=\frac{3}{5}$. As shown in Figure (1), $0<\angle A O B=\angle C O A+\angle C O B<\pi$, then $\cos \angle A O B=\cos (\angle A O C+\angle C O B)=\cos \angle A O C \cos \angle C O B-\sin \angle A O C \sin \angle C O B=$ $0, \therefore \angle A O B=\frac{\pi}{2}$, i.e., $a \cdot b=0, \therefore z^{2}=|c-t a-b|^{2}=c^{2}+t^{2} a^{2}+b^{2}-2 t c \cdot a-2 c$ $\cdot \boldsymbol{b}+2 t a \cdot b=t^{2}-6 t+18=(t-3)^{2}+9 \geqslant 9$, so when $t=3$, $z_{\min }=3$;

As shown in Figure (2), $0<\angle A O B=\angle C O A-\angle C O B<\frac{\pi}{2}$, then $\cos \angle A O B=\cos (\angle A O C$ $\angle C O B)=\cos \angle A O C \cos \angle C O B+\sin \angle A O C \sin \angle C O B=\frac{24}{25}$,
$\therefore \boldsymbol{a} \cdot \boldsymbol{b}=|\boldsymbol{a}||\boldsymbol{b}| \cos \angle A O B=\frac{24}{25}$,
$\therefore z^{2}=|c-t a-b|^{2}=c^{2}+t^{2} a^{2}+b^{2}-2 t c \cdot a-2 c \cdot b+2 t a \cdot b=t^{2}-\frac{102}{25} t+18=$
$\left(t-\frac{51}{25}\right)^{2}+\left(\frac{93}{25}\right)^{2} \geqslant\left(\frac{93}{25}\right)^{2}$, so when $t=\frac{51}{25}$, $z_{\min }=\frac{93}{25}$; In summary, when $t=3$, $z_{\min }=3$.