Let the vectors \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) be such that \(|\bar{a}|=2 ;|\vec{b}|=4\) and \(|\bar{c}|=4\). If the projection of \(\bar{b}\) on \(\bar{a}\) is equal to the projection of \(\bar{c}\) on \(\bar{a}\) and \(\bar{b}\) is perpendicular to \(\bar{c}\), then the value of \(|\bar{a}+\bar{b}-\bar{c}|\) is

\(|\overrightarrow{\mathrm{a}}|=2,|\overline{\mathrm{~b}}|=4,|\overrightarrow{\mathrm{c}}|=4\)
According to the given condition, Projection of \(\overline{\mathrm{b}}\) on \(\overline{\mathrm{a}}=\) projection of \(\overline{\mathrm{c}}\) on \(\overline{\mathrm{a}}\)
\(\begin{aligned}
& \Rightarrow \frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}}{|\overline{\mathrm{a}}|}=\frac{\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}}{|\overline{\mathrm{a}}|} \\
& \Rightarrow \overline{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}=\overline{\mathrm{c}} \cdot \overline{\mathrm{a}} \\
& \Rightarrow(\overline{\mathrm{~b}}-\overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}=0...(i)
\end{aligned}\)
Now, \(|\bar{a}+\bar{b}-\bar{c}|=\sqrt{|\bar{a}+\bar{b}-\bar{c}|^2}\)
\(=\sqrt{|\bar{a}|^2+|\bar{b}-\bar{c}|^2+2 \bar{a} \cdot(\bar{b}-\bar{c})}\)
\(=\sqrt{(2)^2+|\bar{b}-\bar{c}|^2+2(0)}\)
...[From (i)]
\(=\sqrt{4+|\vec{b}-\bar{c}|^2}\)
\(=\sqrt{4+|\bar{b}|^2+|\bar{c}|^2-2(\bar{b} \cdot \bar{c})}\)
\(\begin{aligned}
& =\sqrt{4+4^2+4^2-2(0)} \\
& =\sqrt{36}=6
\end{aligned}\)
...[ \(\bar{b}\) is perpendicular to \(\bar{c}]\)