Prove a Set is Closed Under Addition

Question

How would you prove that the set of all vectors in $\mathbb{R}^3$ whose coordinates $x, y, z,$ satisfy the the equation

$$6x - 2y - 4z =0$$

is closed under vector addition?

I know that the set only contains elements where $x = y = z$, but how would that translate into a proof that the set is closed under addition?

Answer 1

Suppose that two vectors $\vec v=(x_1,y_1,z_1)$ and $\vec w=(x_2,y_2,z_2)$ are both in the set described. Then $$6x_1 - 2y_1 - 4z_1 =0$$ $$6x_2 - 2y_2 - 4z_2 =0$$ and so, by adding these two equations, $$6(x_1+x_2) - 2(y_1+y_2) - 4(z_1+z_2) =0$$ which implies that the vector $\vec v+\vec w=(x_1+x_2,y_1+y_2,z_1+z_2)$ is also in the described set. Thus, since $\vec v$ and $\vec w$ being in the set implies that $\vec v+\vec w$ is also in the set, it is closed under vector addition. $\blacksquare$

Answer 2

suppose that $(x,y,z),(a,b,c)$ satisfy the equation.

Then $(x-2y-4z)+(a-2b-4c)=0$, but then $(x+a)-2(y+b)-4(z+c)=0$.

Answer 3

Let $6x - 2y - 4z =0$ and $6a - 2b - 4c =0$.

We know $(x, y,z) + (a,b,c) = (x+a, y+ b, z+c)$

So what does $6(x+a) - 2(y+b) - 4(z + c)$ equal?