# Prove that two vectors are perpendicular to each other

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?

• Using Cauchy-Schwarz is too strong, the condition $\|x\|\|y\|=0$ means either $x$ or $y$ must be $0$ and this is stronger than just $x$ being orthogonal to $y$. Jun 23, 2021 at 10:26

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 $$\langle x,y\rangle=0$$.

• who downvoted and why? Jun 23, 2021 at 10:33
• How do you know that $x+y$ and $z$ are perpendicular to each other (assumption of pythagorean theorem)? I didn't downvote - just to add ;)) Jun 23, 2021 at 10:33
• @Lucian I mean the pythagorean theorem in its generalized form, as mentioned by OP: $\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$. No orthogonality is assumed for this. Jun 23, 2021 at 10:33
• I've usually heard that called the parallelogram identity
– Alan
Jun 23, 2021 at 10:34
• @Lucian An inner product space, by definition, is a vector space with an inner product that induces a norm
– Alan
Jun 23, 2021 at 10:36

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 $$a=2\langle x,y\rangle$$.

• this answer also assumes that the vector space is over the field of real numbers Jun 23, 2021 at 10:44
• @JustDroppedIn Indeed. I've added that assumption to my answer. Jun 23, 2021 at 10:47