Cauchy-Shwarz inequality in vector analysis

Question

Vectors $x$ and $y$ are related as follows $$\mathbf{x}+\mathbf{y(x \cdot y)}=\mathbf{a}.$$

Show $$\mathbf{(x \cdot y)}^2=\mathbf{\frac{|a|^2-|x|^2}{2+|y|^2}}$$

I think we need to proceed using Cauchy-Shwarz inequality.

$\mathbf{y(x \cdot y)}=\mathbf{a}-\mathbf{x}$

$\mathbf{y(y \cdot x)(y \cdot x)}=(\mathbf{a}-\mathbf{x)(x \cdot y)}$

$\mathbf{y(y \cdot x)^2}=(\mathbf{a}-\mathbf{x)(x \cdot y)}$

Then, I am lost.

Answer 1

We have $a = x + \def\<#1>{\left<#1\right>}\<x,y>y$, hence \begin{align*} |a|^2 &= \<a,a>\\ &= \<x + {\<x,y>y, x+ \<x,y>y}>\\ &= \<x,x> + 2\<x,y>^2 + \<x,y>^2\<y,y>\\ &= |x|^2 + (2 + |y|^2)\<x,y>^2\\ \iff |a|^2 - |x|^2 &=(2 + |y|^2)\<x,y>^2\\ \iff \<x,y>^2 &= \frac{ |a|^2 - |x|^2 }{2 + |y|^2} \end{align*}