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Given a vector $A$ different of zero, and a vector $C$ orthogonal to $A$, both vectors in $\mathbb{R}^3$, prove that there is exactly a vector $B$ such that A$\times$B=C and $A\cdot B=1$.

And i did the following: i proposed $A=(a_1,a_2,a_3)$, $C=(c_1,c_2,c_3)$ and $B=(x,y,z)$ then i used the conditions of the problem and i got a system of linear equations

$a_1c_1+a_2c_2+a_3c_3=0$

From $A\times B=C$, I obtained

$c_1=a_2z-a_3y$

$c_2=a_3x-a_1z$

$c_3=a_1y-a_2z$

And finally

$a_1x+a_2y+a_3z=1$

But i dont know how to conclude that there is a unique solution for this system, can you help with that?

And a last question, my professor recommended us to use Lagrange identity to prove this but i really have no idea of how to achieve this, can you give some hints, please?

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  • $\begingroup$ Can you pimp your question with mathjax? $\endgroup$ Aug 30, 2017 at 7:20
  • $\begingroup$ @xyzt has done it, thks! $\endgroup$
    – Sama
    Aug 30, 2017 at 7:24

2 Answers 2

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Suppose you have two solutions, $B$ and $B'$. Then $A\times B=A\times B'=C$ and therefore $A\times(B-B')=0$. Since $A\neq0$, $B-B'=\lambda A$, for some $\lambda\in\mathbb R$. So$$1=A.B=A.(B'+\lambda A)=A.B'+\lambda\|A\|^2=1+\lambda\|A\|^2$$and therefore $\lambda=0$. This means that $B'=B$.

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You must have $$A\times (A\times B)=A\times C.$$ This is equivalent to $$(A\cdot B)A-(A\cdot A)B=A\times C.$$ We know $A$, $C$ and $A\cdot B$....

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