If $\vec{u}+\vec{v}+\vec{w}=\vec{0}$, then triangle $ABC$ exists with $\vec{AB}=\vec{u},\vec{BC}=\vec{v},\vec{CA}=\vec{w}$ Let $ABC$ be a triangle. Then $\vec{AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}=\vec{AA}=\vec{0}.$ Is the opposite statement that if $\vec{u}+\vec{v}+\vec{w}=\vec{0}$, then triangle $ABC$ exists with $\vec{AB}=\vec{u},\vec{BC}=\vec{v},\vec{CA}=\vec{w}$ also true? Why?
I am really new to vectors and I would be very grateful if someone could explain this to me in detail. Thank you in advance!
The opposite statement is not true. It suffices to consider $A,B,C$ collinear.
Set $$\vec{AB}=\vec{u},\vec{BC}=\vec{v},\vec{CA}=\vec{w}$$
It holds $$\vec{u}+\vec{v}+\vec{w}=\vec{0},$$ but $ABC$ is not a triangle.
Take any point $A$. Define $B$ as $A+\overrightarrow{u}$ and $C$ as $B + \overrightarrow{v}$. Then
$$ C+\overrightarrow{w} = B + \overrightarrow{v}+\overrightarrow{w} = A+\overrightarrow{u}+ \overrightarrow{v}+\overrightarrow{w} = A $$
so triangle $ABC$ indeed satisfies our thesis.