Let $\vec x=(x_1,-2,1,-1)$ and $\vec y=(-2,y_2,-1,-2)$, wich satisfies $\lVert \vec x\rVert=2\lVert \vec y\lVert$. Find the indeterminate values of $x_1$ and $y_1$.
So, assuming $\lVert _\dot{} \lVert =\lVert _\dot{}\lVert_2$, then:
$$\lVert \vec x\rVert:=\Bigg(\sum_{i=1}^{4} x_i^2\Bigg)^{1/2}=\sqrt {x^2_1+(-2)^2+1+(-1)^2}=\sqrt{x^2_1+6}$$ $$\lVert \vec y\rVert:=\Bigg(\sum_{i=1}^{4} y_i^2\Bigg)^{1/2}=\sqrt {(-2)^2+y^2_2+(-1)^2+(-2)^2}=\sqrt{y^2_2+9}$$ Then, $$\sqrt{x^2_1+6}=2\sqrt{y^2_2+9}$$ $$x^2_1+6=4y^2_2+36$$ $$x^2_1=4y^2_2+30$$ $$x_1=\sqrt{4y^2_2+30}$$ Now, can it be said that every $x_1,y_2\in\Bbb R$ satisfies the condition $\lVert \vec x\rVert=2\lVert \vec y\lVert$? 'Cause we can solve for $y_2$ as well: $$x^2_1=4y^2_1+30$$ $$\implies y_2=\frac{\sqrt{x^2_1-30}}{2}, x^2_1\ge30$$ And it follows that $$x^2_1\ge30\implies x_1\le -\sqrt{30} \lor x_1\ge \sqrt{30}$$ And so we get: $x_1\in(-\infty,-\sqrt{30}]\cup[\sqrt{30},\infty)$
Now we got the set of solutions: $S=\{(x_1,y_2)|y_2=\frac{\sqrt{x^2_1-30}}{2},x_1\in(-\infty,-\sqrt{30}]\cup[\sqrt{30},\infty)\}$
Is this the proper way to state it?
Thanks for the feedback.