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$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors such that $|\mathbf{a}|=3$, $|\mathbf{b}|=5,|\mathbf{c}|=7$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are perpendicular to the vectors $\mathbf{b}+\mathbf{c}, \mathbf{c}+\mathbf{a}, \mathbf{a}+\mathbf{b}$ respectively, then $\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})^2-2}=$
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Verified Answer
The correct answer is:
9
We have,
$$
|\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7
$$
$\mathbf{a}$ is perpendicular to $(\mathbf{b}+\mathbf{c})$

and $\mathbf{b}$ is perpendicular to $\mathbf{c}+\mathbf{a}$ So,
$\mathbf{b} \cdot(\mathbf{c}+\mathbf{a})=0$
and $\mathbf{c}$ is perpendicular to $(\mathbf{a}+\mathbf{b})$
Adding Eqs. (i), (ii) and (iii), we get
Now, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2$
$$
\begin{aligned}
& =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\
& =(3)^2+(5)^2+(7)^2+0=83
\end{aligned}
$$
Now, $\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})^2-2}=\sqrt{83-2}=\sqrt{81}=9$
$$
|\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7
$$
$\mathbf{a}$ is perpendicular to $(\mathbf{b}+\mathbf{c})$

and $\mathbf{b}$ is perpendicular to $\mathbf{c}+\mathbf{a}$ So,
$\mathbf{b} \cdot(\mathbf{c}+\mathbf{a})=0$

and $\mathbf{c}$ is perpendicular to $(\mathbf{a}+\mathbf{b})$

Adding Eqs. (i), (ii) and (iii), we get

Now, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2$
$$
\begin{aligned}
& =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\
& =(3)^2+(5)^2+(7)^2+0=83
\end{aligned}
$$
Now, $\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})^2-2}=\sqrt{83-2}=\sqrt{81}=9$
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