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I try it graphically but I don't think its the correct one.enter image description here

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    $\begingroup$ $$(x+y)(x-y)=x^2-y^2$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 12 at 13:25
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    $\begingroup$ Have you tried computing their scalar(/dot/inner) product? $\endgroup$ – StackTD Jul 12 at 13:26
  • $\begingroup$ Remember, this isn't necessarily $\mathbb R^2$, but you can definitely build intuition in $\mathbb R^2$. Note that if $x-y=0$, $x=y$. (Likewise, $x+y=0\to x=-y$). This is a issue, why? Moreover, what vector to scalar function allows you to prove perpendicularity easily? $\endgroup$ – Don Thousand Jul 12 at 13:26
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    $\begingroup$ use properties of the dot product; e.g., commutative, distributive, and if $x$ has unit length then $x\cdot x=1$ $\endgroup$ – J. W. Tanner Jul 12 at 13:29
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    $\begingroup$ write both unit vectors in component form and show that the dot product of $(x-y) \cdot (x+y)=0$ $\endgroup$ – Vasya Jul 12 at 13:30
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Using properties of the dot product, $(x+y)\cdot(x-y)=x\cdot x - x\cdot y + y \cdot x - y \cdot y=x\cdot x - y\cdot y. $

Furthermore, if $x$ and $y$ are unit vectors, then $x\cdot x=y\cdot y=1,$ so $(x+y)\cdot(x-y)=0. $

What does that mean?

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Let $x=(x_1,x_2,..,x_n)$ and $y=(y_1,y_2,..,y_n)$ then, $\sum_{k=1}^{k=n}{x_k^2}=1$, $\;$ $\sum_{k=1}^{k=n}{y_k^2}=1$, $\;$ $x-y=(x_1-y_1,x_2-y_2,..,x_n-y_n)$ and $x+y=(x_1+y_1,x_2+y_2,..,x_n+y_n)$. With dot product, $0=\sum_{k=1}^{k=n}{(x_k^2-y_k^2)}=(x-y).(x+y)=|x||y|cos\alpha=cos\alpha$ (here $\alpha$ is the angle between x and y) thus, $\alpha=\pi/2$

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  • $\begingroup$ That is not quite right: You assume that their dot product is $0$ and then show that this implies that the angle between them is $\frac{\pi}{2}$. What you instead need to use is that they are both unit vectors. $\endgroup$ – asdf Jul 12 at 13:55
  • $\begingroup$ Maybe now it is more clear. $\endgroup$ – Taha Direk Jul 12 at 13:59

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