Show that if |A-B|=|A+B|, then A and B must be orthogonal.

Question

Is there a better way to show it?

My attempt:

Suppose A & B $ \neq 0$

Squaring both sides will result in $ A^2-2AB+B^2=A^2+2AB+B^2$

$4AB=0 \iff A.B=0 $

A&B must be orthogonal

Answer 1

Idea is right, but you don't need suppose A & B ≠ 0 which makes it sound like a proof by contradiction. Besides, the notation $A \,\&\, B$ for the dot product is non-standard. More common:

• $|A-B|^2= \langle A-B, A-B \rangle= |A|^2 - 2\,\langle A,B\rangle+|B|^2\,$, then $\;\cdots\;$ $\langle A, B \rangle = 0 \iff A \perp B$

• $|A-B|^2= (A-B) \cdot (A-B)= |A|^2 - 2\,A\cdot B+|B|^2\,$, then $\;\cdots\;$ $A \cdot B = 0 \iff A \perp B$

Answer 2

A little Geometry :

Draw two vectors from (0,0), not colinear:

Call them $\vec{a}$ and $\vec{b}$.

Label $A, B$ the endpoints of the two vectors resp.

Vector addition:

$ \vec{c} := \vec{a} + \vec{b}$.

Let the endpoint of $\vec{c}$ be $C$.

Consider the parallelogram $(0,0), A,C,B$.

1) $\vec{c}$ is along the diagonal.

2) Vector $ \vec{d}$ : = $\vec{b} - \vec{a}$ is along the other diagonal.

Given:

$|\vec{c}| = |\vec{d}|$ =

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

Hence the two diagonals of the parallelogram have equal length.

$\rightarrow$ the parallelogram is a rectangle, I.e.

$\vec{a}$ is perpendicular to $\vec{b}$.