Mathematics - Vector Question with Solution | TestHub

$\begin{array}{l}\because |\vec{a}|=|\vec{b}|=|\vec{c}|=1\\ |\vec{a}-\vec{b}{|}^2+|\vec{b}-\vec{c}{|}^2+|\vec{c}-\vec{a}{|}^2=9\\ \Rightarrow 2\left(|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2\right)-2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=9\\ \Rightarrow \vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=\frac{-3}{2}\end{array}$ Also, $\begin{array}{l}|\vec{a}+\vec{b}+\vec{c}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})\\ =1+1+1+2\times \left(-\frac{3}{2}\right)=0\\ \Rightarrow \vec{a}+\vec{b}+\vec{c}=0\Rightarrow (\vec{b}+\vec{c})=-\vec{a}\\ \therefore |2\vec{a}+5(\vec{b}+\vec{c})|=|2\vec{a}-5\vec{a}|=|-3\vec{a}|=3\end{array}$