Prove that there exists a unique $ v \in \mathbb{R}^3 $ such that $ u \times v = w $ and $ u \cdot v = 1 $ Let $ u, w \in \mathbb{R}^3$ be such that $ u \neq 0 $ and $ w $ is orthogonal to $ u $. Prove that there exists a unique $ v \in \mathbb{R}^3 $ such that $ u \times v = w $ and $ u \cdot v = 1 $
My attempt:
Let $u = (u_1 , u_2 , u_3)$, $w=(w_1 , w_2, w_3)$ and $v=(v_1 , v_2, v_3)$. I need to find $v_1$, $v_2$ y $v_3$ such that $v $ satisfies the conditions of the problem. Considering the equation $u \times v = w $ and calculating $v_1$, $v_2$ and $v_3$ we get that $v_1 = - \left ( \frac{w_2}{u_3} + \frac{u_{1}(w_{1}u_{1} - w_{2}u_{2})} {u_{1}u_{2} - u_{1}u_{3}}\right)$, $v_2 = \frac {w_1}{u_3} - \frac {u_2}{u_3} (\frac {w_{1}u_{1} + w_{2}u_{2}}{u_{1}u_{2} - u_{1}u_{3}})$ y $v_{3} = \frac {-(w_{1}u_{1} + w_{2}u_{2})}{u_{1}u_{2} - u_{1}u_{3}}$. Now, I also need that $v$ satisfy the equation $u \cdot v = 1$. I have that $ u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}= 1$, but I have not concluded anything. I need some help to do this. On the other hand, I suppose that for the uniqueness of $v$ it is better to show that the matrix of coefficients obtained in the previous step has a unique solution. Or perhaps it is better to show that given another vector $v'$ such that $u \times v' = w$ and $u \cdot v' = 1$, we must have $v' = v$.
 A: If $v_1,\,v_2$ both work then $u\times(v_1-v_2)=0,\,u\cdot(v_1-v_2)=0$, and the only choice of $v_1-v_2$ both parallel and perpendicular to $u$ is the zero vector. So uniqueness is the easy part.
In terms of $\hat{u}:=\tfrac{u}{|u|},\,\hat{w}:=\tfrac{w}{|w|}$ define $x:=\hat{u}\times\hat{w}$ so $(\hat{u},\,\hat{w},\,x)$ is a right-handed $3$-tuple of orthonormal vectors. The Ansatz $v=Ax+Bu$ simplifies the problem to$$A|u|\underbrace{\hat{u}\times x}_{-\hat{w}}=|w|\hat{w},\,B|u|^2=1$$i.e. $A=-\tfrac{|w|}{|u|},\,B=\frac{1}{|u|^2}$. Hence$$v=-\frac{|w|}{|u|}\frac{u}{|u|}\times\frac{w}{|w|}+\frac{u}{|u|^2}=\frac{u-|w|u\times w}{|u|^2}.$$
A: Think geometrically $u \times v = w$ means $w$ is the normal vector to $u,v$. So if we start with $w$, this has a normal plane, in which $u$ lies. A plane is two dimensional so we need to find another basis element besides $u$ which has some fixed angle to $u$ (this is what $u \cdot v = 1$ means). It's clear geometrically there can be but one such vector that such that $\{u,v\}$ span this normal plane and $(u,v)$ has the right orientation.
A: Here's a less geometric, more linear algebraic approach.
Let $U = u^\bot \subseteq \mathbb{R}^3$ and let $f:\mathbb{R}^3\rightarrow U\oplus \mathbb{R}$ be the function $f(v) = (u\times v, u\cdot v)$.  Then $f$ is a linear map between these two vector spaces.
Moreover, the kernel of $f$ is consists of those $v$ which are simultaneously parallel to $u$ (because of the first slot) and perpendicular to $u$ (because of the second slot).  Since $u\neq 0$, the only such vector is $0$, so the kernel of $f$ is trivial.
So, $f$ is an injective linear map between two vector spaces of the same (finite) dimension, so it is an isomorphism.  In particular, $f^{-1}$ exists and the $v$ you want is uniquely defined as $f^{-1}(w,1)$.
A: You are looking for a $v$ in the plane perpendicular to $w$, such that the orthogonal projection of $v$ to $u$ is a fixed vector (this is what fixing $u\cdot v$ means). Thus $v$'s endpoint lies on a predetermined line orthogonal to $u$. The final condition $u\times v=w$ specifies the oriented area of the triangle formed by $u$, $v$ and the origin. By varying $v$'s endpoint in this line any oriented are can be achieved, and in a unique way. Thus the $v$ you are looking for always exists and is unique.
A: If you want to use coordinates, you can set the coordinates system so that $w=\langle 0,0,w_3 \rangle$. Since $u$ and $v$ are perpendicular to $w$, they can be written as $u=\langle u_1,u_2,0 \rangle$ and $v=\langle v_1,v_2,0 \rangle$. Then $v$ has to be solution of
$$u_1v_1+u_2v_2=1 \hskip 4mm \text{ and }\hskip 4mm u_1v_2-u_2v_1=w_3$$
This is a linear system with unknowns $v_1$ and $v_2$. Since its coefficient matrix $A=\begin{bmatrix} u_1 & u_2\\ -u_2 & u_1  \end{bmatrix}$ has $u_1^2+u_2^2=||u||^2>0$ for determinant, the system has a unique solution.
A: A vector in spherical coordinates uniquely determined by $(\rho,\varphi,\theta)$. (norm of vector, angle with $+uv$-plane and angle with $+u$-axis). We know that $\varphi=0$ since it is in $vu$-plane. $\theta=\pm \tan^{-1}|w|$ (one sign is acceptable and characterized by $u\times v$.) since $|v||u|\cos\theta=1$ and $|w|=|v||u|\sin\theta$. So $\tan\theta=|w|$. Because $\theta$ is uniquely determined then $\rho=|v|$ is unique from $|v||u|\cos\theta=1$.
