# Vector Orthogonality and Length

The problem statement, all given variables and data

Let $\textbf{a}$, $\textbf{b}$ be two vectors in $\mathbb{R}^n$. If $\textbf{a} + \textbf{b}$ and $\textbf{a} - \textbf{b}$ are orthogonal, then show that $||\textbf{a}|| = ||\textbf{b}||$.

Relevant equations

For two vectors to be orthogonal, their dot product must equal to $0$. Vector length is acquired by taking the square of each element in the vector then taking the root of their addition.

Attempt at a solution

I basically wrote down the formula: $(\textbf{a} + \textbf{b}) . (\textbf{a} - \textbf{b}) = 0$.

Also, I got a set of two example vectors which satisfy this equation. Vectors $(1, 0, 0)$ and $(0, 1, 0)$ both have the same length and are orthogonal to each other. Their dot product is also $0$. However, how do I provide a general proof to this statement? I don't think I can just give an example and say "oh look it works here so it must work everywhere".

• $3$ identical answers in the first minutes. $1$ more please :p Aug 12, 2014 at 13:57

$$(a+b)\cdot(a-b)=0$$ $$a\cdot a - a \cdot b + b \cdot a -b\cdot b=0$$ $$a\cdot a =b\cdot b$$ $$||a||^2=||b||^2$$ $$||a||=||b||$$
$$0=(a+b)\cdot (a-b)=a^2+a\cdot b-a\cdot -b^2=a^2-b^2\iff a^2=b^2\iff \|a\|^2=\|b\|^2$$
Using the associative and commutative properties of the dot product and the formula $a \cdot a = ||a||^2$, we have $$(a+b) \cdot (a-b) = (a+b) \cdot a - (a+b) \cdot b = a \cdot a + b \cdot a - a \cdot b - b \cdot b = ||a||^2 + a \cdot b - a \cdot b - ||b||^2 = ||a||^2 - ||b||^2.$$ So $$(a+b) \cdot (a-b) = 0$$ if and only if $$||a||^2 - ||b||^2 = 0$$ if and only if $$||a||^2 = ||b||^2$$ if and only if $$||a|| = ||b||,$$ which is the desired result.