Let $v = (-1,0,2)$ and $u = (3,1,-2)$ so we need to find $x = (x_1,x_2,x_3)$ such that $v \cdot x = 0$ and $u \cdot x = 0$. This gives us the set of linear equations
$$ -x_1+2x_3 = 0$$
$$ 3x_1 + x_2 – 2x_3 = 0$$
which, through elementary row operations gets us the two linear equations
$$ x_1-2x_3 = 0$$ $$x_2 + 4x_3 = 0 $$
so $x_1 -2x_3 = x_2 + 4x_3$ which simplifies to $x_1 = x_2 + 6x_3$. so the solution set would be
{$(x_2+6x_3, x_2, x_3)$ | $x_3,x_2\in \mathbb R$}
But the answer is {$r(2,-1,4)$ | $r \in R$}. How do i get this/did I make an error?
Best Answer
Your mistake was to assume that $x_2$ and $x_3$ are indepedent, what is not true, since you have got the equation $x_2+4x_3=0$. To get the right result, just chose one variable to be parametrized and using both equations you got, find the relation of the other two variables to this parameter.