Prove that two vectors are perpendicular to each other

Question

Let's consider inner product space with vectors $x, y, z$ which satisfies:

$$\|x+y+z\|^2 = 14$$

$$\|x+y-z\|^2 = 2$$

$$\|x-y+z\|^2 = 6$$

$$\|x-y-z\|^2 = 10$$

I want to prove that $x$ is perpendicular to $y$.

My work so far

In other words we want to prove that $\langle x, y \rangle = 0$.

My first idea was to use Cauchy Schwarz inequality $ | \langle x , y \rangle |\le \|x\| \|y\|$ and to show that our conditions force that $\|x\|\cdot\|y\| = 0$. However I didn't manage to do anything sensible. Also I tried to somehow prove that under this conditions our norm has bo inducted by inner product - in other words we have that: $$2(\|x\|^2 + \|y\|^2) = \|x+y\|^2 + \|x-y\|^2$$

but also I didn't end up with something rational. Could you please give me a hand, what's the correct approach to this problem?

Answer 1

I assume you work with a real and not a complex vector space. Is that true?

Add the first and second one: by the parallelogram identity you get $$2(\|x+y\|^2+\|z\|^2)=16$$ so $\|x+y\|^2+\|z\|^2=8$. Now add the third and the fourth one and in the same way you get $\|x-y\|^2+\|z\|^2=8$. So $\|x+y\|^2+\|z\|^2=\|x-y\|^2+\|z\|^2$, so $$\|x-y\|^2=\|x+y\|^2$$ so $\|x\|^2+\|y\|^2-2\langle x,y\rangle=\|x\|^2+\|y\|^2+2\langle x,y\rangle$, so $\langle x,y\rangle=0$.

Answer 2

Assuming that you are working over $\Bbb R$, let:

• $s=\|x\|^2+\|y\|^2+\|z\|^2$;
• $a=2\langle x,y\rangle$;
• $b=2\langle x,z\rangle$;
• $c=2\langle y,z\rangle$.

Then those four equalities tell you that$$\left\{\begin{array}{l}s+a+b+c=14\\s+a-b-c=2\\s-a+b-c=6\\s-a-b+c=10.\end{array}\right.$$In particular,$$\begin{split}(s+a+b+c)+(s+a-b-c)-(s-a+b-c)-(s-a-b+c) &=14+2-6-10 \\ &=0 \end{split}$$In other words, $4a=0$. But $a=2\langle x,y\rangle$.