Question

Does A = {(x,y,z) ∈R3 : x + 3y + 2z = 0} build a set in R3

Answer 1

We need to show that A is a subspace of $\R^3$

Since $(0,0,0)\in \R^3$ and satisfies the condition on A, then A is not empty.

Let $a=(a_1,a_2,a_3), b=(b_1,b_2,b_3)\in A, \alpha\in \R | a_1+3a_2+2a_3=0,b_1+3b_2+2b_3=0$

We shall show that $\alpha(a+b)\in A$

$\alpha(a+b)=(\alpha(a_1+b_1),(\alpha(a_2+b_2),(\alpha(a_3+b_3))\\ (\alpha(a_1+b_1)+3(\alpha(a_2+b_2)+2(\alpha(a_3+b_3)=\alpha(a_1+3a_2+2a_3)+\alpha(b_1+3b_2+2b_3)=0$

$\implies \alpha(a+b)\in A$

Hence A is a subspace of $\R^3$

Therefore, A build a set in $\R^3$