Let $\vec{a},\vec{b},$ and $\vec{c}$ be vectors such that $|\vec{a}|=3,|\vec{b}|=4,$ and $|\vec{c}|=5$. If $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, then the value of $\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}$ is:

Solution

1. Consider $$\begin{array}{rl}|\vec{a}+\vec{b}+\vec{c}{|}^2& =|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})\\ & =0\end{array}$$.
2. Substitute the magnitudes: $3^2+4^2+5^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0$.
3. Calculate the sum of squares: $9 + 16 + 25 = 50$.
4. Thus, $50+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0$, so the sum is $-25$.
5. Hence the answer is (A).