How to prove $\vec{a} \perp \vec{b}$ when $|\vec{a}-\vec{b}|=|\vec{a}+\vec{b}|$?

Question

I have tried to show that $cos(\theta)=0$ by substituting what is given into $(\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b})=|\vec{a}-\vec{b}||\vec{a}+\vec{b}|cos(\theta)$ but after simplifying i get stuck at

$cos(\theta)=\frac{(\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b})}{|\vec{a}+\vec{b}||\vec{a}+\vec{b}|}=\frac{(\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b})}{(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})}$

Answer 1

$|\vec a-\vec b|^2=|\vec a+\vec b|^2\implies(\vec a-\vec b)\cdot(\vec a-\vec b)=(\vec a+\vec b)\cdot(\vec a+\vec b)\implies \vec a\cdot \vec b=0$

Answer 2

Square the equality,

$$|\vec{a}-\vec{b}|=|\vec{a}+\vec{b}|$$

to get

$$-\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{b}$$

which yields

$$\vec{a}\cdot\vec{b}=0$$

Thus, $\vec{a} \perp \vec{b}$

Answer 3

Just for the record: Using the the parallelogram identity $$2|a|^2+2|b|^2=|a+b|^2+|a-b|^2$$ we get $$|a|^2+|b|^2=|a+b|^2.$$ From Pythagoras we arrive in $a\perp b$.

Answer 4

A bit of geometry in $\mathbb{R^2}$:

Consider the triangle formed by $-\vec b$, from origin along negative $x$-axis to point $A$; $\vec b$, from origin to point $B$; $\vec a$, from origin to point $C$.

In $\triangle A,B,C$ :

Length of side $\overline {BC}= |\vec a -\vec b|$;

Length of side $\overline {AC}= $

$|-\vec b -\vec a|=|\vec b +\vec a|.$

$\vec a$ is along a median on side $AB$.

$\overline {AC}=\overline {BC}$ $\iff$ median is

perpendicular bisector, i.e. $\vec a \perp \vec b$.