# If a convex set in $R^3$ contains the three given vectors, does it also contain the zero vector?

I tried reasoning it out below. I am not sure if it works or not. Please, see if it needs to be redone and/or mistakes fixed.

Let $S$ be a convex set containing $(1, 2, 3), (-1, -3, 4), (0, 1, -7).$ Then $a_1(1, 2, 3) + a_2(-1, -3, 4) + a_3(0, 1, -7) \in S$ where $\sum a_i = 1$ and $a_i \in [0, 1].$ If we can show that this linear combination is $(0, 0, 0)$ for $\{a_1, a_2, a_3\}$ with the given properties, we are done. Consider the system of equations below:

$$\begin{cases} a_1 -a_2+0=0 \\[2ex] 2a_1 -3a_2+a_3=0 \\[2ex] 3a_1+ 4a_2 -7a_3=0 \end{cases}$$

Then $a_1 = a_2$. Substituting $a_1$ for $a_2$ in the second equation we get $-a_1 = -a_3.$ So then $a_1 = a_2 = a_3.$ But there are no such $a_i$ with the given properties.

edit: argh, I just realized I am a dumbo. Thanks, everyone.

• Don't each of those vectors start at the origin? – IAmNoOne Nov 29 '15 at 23:55
• Doesn't $a_1 = a_2 = a_3 = \frac 1 3$ work? – Alvin Jin Nov 30 '15 at 0:00

You got $a_1 = a_2 = a_3$. Since their sum is $1$, all of them are $1/3$. And: $$\frac{1}{3}((1,2,3)+(-1,-3,4)+(0,1,-7)) = \frac{1}{3}(0,0,0) = (0,0,0).$$Since the zero vector is a convex combination of the above vectors and the set is convex, the zero vector is also there. No problems.