3. Given that $\boldsymbol{a}, \boldsymbol{b}$ are unit vectors, $|3 \boldsymbol{a}+4 \boldsymbol{b}|=|4 \boldsymbol{a}-3 \boldsymbol{b}|$, if $|\boldsymbol{c}|=2$, then the maximum value of $|\boldsymbol{a}+\boldsymbol{b}-\boldsymbol{c}|$ is
Solve:
$|3 \boldsymbol{a}+4 \boldsymbol{b}|=|4 \boldsymbol{a}-3 \boldsymbol{b}| \Rightarrow \boldsymbol{a} \cdot \boldsymbol{b}=0 \Rightarrow \boldsymbol{a} \perp \boldsymbol{b}$. Let's assume $\boldsymbol{a}=(1,0), \boldsymbol{b}=(0,1)$, so $|\boldsymbol{a}+\boldsymbol{b}-\boldsymbol{c}| \leqslant|\boldsymbol{a}+\boldsymbol{b}|+|\boldsymbol{c}|=\sqrt{2}+2$.