Vectors of same length, perpendicularity

Question

How to algebraically show that if two vectors i.e. $\vec a$ and $\vec b$ have the same length then $\vec a+\vec b$ vector is perpendicular to $\vec a-\vec b$?

I get this graphically but cannot put it algebraically.

Answer 1

For vectors $a$ and $b$ to have the same length means that $\langle a,a\rangle=\langle b,b\rangle$. Now use bilinearity of the scalar product: $$\langle a+b,a-b\rangle=\langle a,a\rangle-\langle a,b\rangle+\langle b,a\rangle-\langle b,b\rangle=0$$.

Answer 2

If $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular, then $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b})=0$. You can desenvolve that to conclude $|\vec{a}|=|\vec{b}|$.

Answer 3

Let $V$ be a real inner product space and $a, b\in V$ with $\|a\|=\|b\|$. Then \begin{align} \langle a+b, a-b\rangle &= \langle a,a\rangle -\langle a,b\rangle +\langle b,a\rangle -\langle b,b\rangle\\ &= \|a\| - 0 - \|b\|\\ &= 0. \end{align} It follows that $a\perp b$.

Answer 4

Hint: You may use

(1) $(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})=0$

(2) for any vector $\vec{v}$, $|\vec{v}|^2=\vec{v}\cdot\vec{v}$

Answer 5

Consider \begin{align} (\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b}) &= \mathbf{a}\cdot (\mathbf{a}-\mathbf{b}) + \mathbf{b}\cdot (\mathbf{a}-\mathbf{b})\\ &= \mathbf{a}\cdot \mathbf{a} - \mathbf{a}\cdot \mathbf{b} + \mathbf{b}\cdot \mathbf{a} - \mathbf{b}\cdot \mathbf{b}\\ &= \mathbf{a}\cdot \mathbf{a} - \mathbf{a}\cdot \mathbf{b} + \mathbf{a}\cdot \mathbf{b} - \mathbf{b}\cdot \mathbf{b}\\ &= \mathbf{a}\cdot \mathbf{a} - \mathbf{b}\cdot \mathbf{b} \end{align}

which we get from bilinearity and symmetry of the inner product (for a real vector space, similar for complex spaces). But $\lVert \mathbf{a} \rVert = \lVert \mathbf{b} \rVert$ implies $\mathbf{a}\cdot \mathbf{a} = \mathbf{b}\cdot \mathbf{b}$ and thus

$$ (\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b}) = 0 \, \text{,} $$ so the vectors $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ are perpendicular (orthogonal).