If \(|\bar{a}|=3,|\bar{b}|=4,|\bar{a}-\bar{b}|=5\), then \(|\bar{a}+\bar{b}|=\)

\(
\begin{aligned}
& |\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2=|\overline{\mathrm{a}}-\overline{\mathrm{b}}|^2+4 \cdot \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \\
& \text { Now }|\overline{\mathrm{a}}-\overline{\mathrm{b}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \\
& \therefore(5)^2=(3)^2+(4)^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=0
\end{aligned}
\)
Substituting in (1), we get
\(
|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2=|\overline{\mathrm{a}}-\overline{\mathrm{b}}|^2 \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|=5
\)