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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.

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  • $\begingroup$ I think the best way is to start with the RHS of the equation you want to prove. Simplify it as much as you can. $\endgroup$ – Friedrich Philipp Mar 7 '16 at 15:22
  • $\begingroup$ @FriedrichPhilipp, I prefer avoiding this method.Vectors have a special property of not being able to get divided and I do not think they obey the law for sum of squares. We are dealing with vectors here. $\endgroup$ – Tosh Mar 7 '16 at 15:34
  • $\begingroup$ martini below exactly did what I suggested. And I don't know what you are talking about here. $\endgroup$ – Friedrich Philipp Mar 7 '16 at 16:03
  • $\begingroup$ @FriedrichPhilipp, I thought that difference of 2 squares cannot be applied in vectors since vectors cannot be factorised. Sorry for misunderstanding. $\endgroup$ – Tosh Mar 7 '16 at 16:06
  • $\begingroup$ First, you don't mean "fatorized" but "multiplied", and, second, you have the squares of the lengths of the vectors here - not the squares of the vectors themselves. $\endgroup$ – Friedrich Philipp Mar 7 '16 at 16:43
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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*}

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