There's this statement in one of my books that says,
In general, for any non-zero vectors $\overrightarrow{a}$, $\overrightarrow{b}$, and $\overrightarrow{c}$, one may note thst although $\overrightarrow{a}+ \overrightarrow{b}+\overrightarrow{c}=0$, they may not always represent the sides of a triangle.
This seems confusing. If I take three vectors, and consider just two of them, I'll get a resultant vector, which can only be cancelled out by another vector parallel to it in the opposite direction, implying that all three have to lie on a triangle. That's the general consensus on three non-zero vectors giving zero resultant that I know of, too.
Is there a case I'm missing here?
Best Answer
OP: “@Yuval, that explains it nicely. Wonder why the book didn't specify this, because it seemed almost intentionally vague, like there were lots of cases left out... are you sure there aren't any more cases?”
Suppose that $\mathbf a,\mathbf b,\mathbf c$ aren't collinear (i.e., aren't parallel).
Then, since $\mathbf a,\mathbf b,\mathbf c$ are nonzero and sum to $\mathbf 0,$ no pair is collinear; in particular, $\mathbf a \nparallel \mathbf b.$
Thus, $0^{\circ}<\text{angle between }\mathbf a \:\&\: \mathbf b <180^{\circ}.$
Since $\mathbf a+\mathbf b+\mathbf c=\mathbf 0,$ the three vectors form a triangle.
Therefore, if the three vectors don't form a triangle, then they must be collinear. In this case, the three vectors in fact form a degenerate triangle.