# Let $x$ and $y$ be unit vectors in $\mathbb{R}^n$, Such that $x+y$ and $x-y$ are non-zero. Show that $x+y$ and $x-y$ are perpendicular.

I try it graphically but I don't think its the correct one.

• $$(x+y)(x-y)=x^2-y^2$$ – Dr. Sonnhard Graubner Jul 12 at 13:25
• Have you tried computing their scalar(/dot/inner) product? – StackTD Jul 12 at 13:26
• 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? – Don Thousand Jul 12 at 13:26
• use properties of the dot product; e.g., commutative, distributive, and if $x$ has unit length then $x\cdot x=1$ – J. W. Tanner Jul 12 at 13:29
• write both unit vectors in component form and show that the dot product of $(x-y) \cdot (x+y)=0$ – Vasya Jul 12 at 13:30

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 $$(x+y)\cdot(x-y)=0.$$
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, $$\alpha=\pi/2$$
• 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. – asdf Jul 12 at 13:55