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$\bar{a}, \bar{b}, \bar{c}$ are vectors such that $|\bar{a}|=5,|\bar{b}|=4,|\bar{c}|=3$ and each is perpendicular to the sum of the other two, then $|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=$
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The correct answer is:
50
We have $\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}}+\overline{\mathrm{c}})=0 \cdot \overline{\mathrm{b}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})=0$ and $\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})=0$
$$
\begin{aligned}
\therefore \quad & \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0 \\
& \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}=0 \\
& \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0
\end{aligned}
$$
From (1), (2) and (3), we get
$$
\begin{aligned}
& 2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=0 \\
& \text { Now }|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\
& \therefore|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=(5)^2+(4)^2+(3)^2+2(0)=50
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad & \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0 \\
& \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}=0 \\
& \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0
\end{aligned}
$$
From (1), (2) and (3), we get
$$
\begin{aligned}
& 2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=0 \\
& \text { Now }|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\
& \therefore|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=(5)^2+(4)^2+(3)^2+2(0)=50
\end{aligned}
$$
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