Showing that vectors x and w are orthoganal.

Question

If $\mathbf{u}$ and $\mathbf{v}$ vectors in $\mathbb{R}^3$. Let $a=|\mathbf{u}|$ and $b=|\mathbf{v}|$. Given that $\mathbf{w} = b\mathbf{u} + a\mathbf{v}$ and $\mathbf{x}= b\mathbf{u}-a\mathbf{v}$ assuming $a, \mathbf{v}, \mathbf{w}, \mathbf{x} \ne 0$. Show that $\mathbf{w}$ and $\mathbf{x}$ are orthogonal.

How would one start this type of question?

Answer 1

\begin{align*} \mathbf{w} \cdot \mathbf{x} &= (b\mathbf{u} + a\mathbf{v}) \cdot (b\mathbf{u}-a\mathbf{v}) \\ &= b^2 \mathbf{u} \cdot \mathbf{u} -ba \mathbf{u} \cdot \mathbf{v}+ ab \mathbf{v} \cdot \mathbf{u}- a^2 \mathbf{v} \cdot \mathbf{u} \\ &= b^2 |\mathbf{u}|^2-a^2 |\mathbf{v}|^2 \\ &= |\mathbf{v}|^2 |\mathbf{u}|^2-|\mathbf{u}|^2 |\mathbf{v}|^2 \\ &= 0 \end{align*}

Answer 2

Hint: Use the properties of an inner product over a real inner product space and a norm arising from an inner product on $\langle bu + av \ , \ bu -av\rangle$.

\begin{align*}\langle bu + av\ , bu - av\rangle&= \langle bu\ , bu-av\rangle + \langle av\ , bu - av\rangle \\&= \langle bu\ , bu \rangle - \langle bu\ , av \rangle + \langle av\ , bu\rangle - \langle av\ , av \rangle \\&= b^2\langle u\ ,u\rangle - a^2\langle v\ ,v\rangle \\& = ||v||^2\langle u\ ,u\rangle - ||u||^2\langle v\ ,v\rangle\\&= ||v||^2\cdot||u||^2 - ||u||^2\cdot||v||^2 \\&= 0\end{align*}