Given two vectors $\vec v, \vec u \ne 0$ and $\alpha$ scalar...

Question

The question is:

Given two vectors $\vec v, \vec u \ne 0$ and $\alpha$ scalar.
is there a vector $\vec w$ such that $\vec u \times \vec w = \vec v$ and $\vec u \cdot \vec w = \alpha $?
In cases that there isn't such a vector $\vec w$, explain why.
In cases that there exists $\vec w$, find it.

My Work:
In order to get some idea on how to do the proof, I tried to take a simple example with coordinations:
$\vec u = (1,1,1)$, $\vec w = (1,0,1)$, then $\alpha = 2$, and $\vec v = \begin{vmatrix}i&j&k\\1&1&1\\1&0&1\end{vmatrix}=(1,0,-1)$
now the question is if I change $\alpha$ or $\vec v$ or $\vec u$, will I be able to find a $\vec w$?
So let's try $\alpha = 3$, now $\vec w$ can be $(1,1,1)$ or $(2,0,1$) ... (for the dot product).
but in the cross product, I can notice that $\vec w$ cannot be equal to $\vec u$ otherwise we get $\vec v=0$.
so I need $\vec v = (1,0,-1)=\begin{vmatrix}i&j&k\\1&1&1\\w_1&w_2&w_3\end{vmatrix}=(w_3-w_2,w_1-w_3,w_2-w_1)$ So I need to find $w_1+w_2+w_3=3$.
But I really just didn't continue here because even if I found it, I don't see myself reaching the proof or atleast getting an idea of how to generalize things, and thought that I'm in somewhat wrong direction or I'm not seeing something important.

After that I tried to use $\vec u, \vec v$ to express $\vec w$ (Just because that's the only way to generalize a vector $\vec w$).
So if I choose lets say: $\vec w = \vec u - \vec v$ I would get $\vec v = \vec u \times (\vec u - \vec v)=\vec u\times \vec u-\vec u \times\vec v = -\vec u \times \vec v$. and again I couldn't find out how to make it equal to $\vec v$, and on top of that I would need to check that it works with the scalar $\alpha$ and the dot product, so I'm a little lost.

End of my work

I would really appreciate any hints or feedback about my two approaches and if they will bring me somewhere, and any pushes in the right direction and help will be awesome, thanks in advance to everyone.

Answer 1

As said in the comments by DonAntonio, a necessary condition for the equation $$\vec u\times \vec{w}=\vec v$$ is that $\vec u,\vec v$ are perpendicular, since $\vec u\times\vec w$ is perpendicular to both $\vec u,\vec w$. It's not that the equation doesn't make sense, but if $\vec u,\vec v$ are not perpendicular, then the equation simply doesn't have a solution, since the existence of a solution $$\vec u\times\vec w=\vec v$$ implies that $$\vec u\cdot\vec v=\vec u\cdot(\vec u\times\vec w)=0$$ that is, $\vec u,\vec v$ are perpendicular.

Now, if $\vec u,\vec v\neq0$ are perpendicular, then the three vectors $$\vec u,\vec v,\vec u\times\vec v$$are three perpendicular vectors in $\mathbb R^3$, therefore they form a basis for $\mathbb R^3$ (in general, in a vector space $V$ with a scalar product $\cdot$, a family of $n=\dim V$ non zero, pairwise orthogonal vectors form a basis for $V$). So, a solution $\vec w$ can be expressed as a unique linear combination $$\vec w=\delta\vec u+\beta\vec v+\gamma(\vec u\times\vec v)$$ then $$\vec u\times\vec w=\delta\underbrace{(\vec u\times\vec u)}_{=0}+\beta(\vec u\times\vec v)+\gamma \vec u\times(\vec u\times\vec v)=\star$$Now, it is a property of the vector product that $\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c$. In this case, set $\vec a=\vec b=\vec u$ and $\vec c=\vec v$ and use that $\vec u\cdot\vec v=0$, so you get $$\vec u\times(\vec u\times\vec v)=-\|u\|^2\vec v$$ so we conclude that $$\star=-\|u\|^2\gamma\vec v+\beta(\vec u\times\vec v)$$ which is equal to $\vec v$ only for $\beta=0$ and $\gamma=-\frac{1}{\|u\|^2}$. So $\vec w$ must be in the form $$\delta\vec u-\frac{1}{\|u\|^2}(\vec u\times\vec v)\rightarrow\vec u\cdot\vec w=\delta\|u\|^2-\frac{1}{\|u\|^2}\underbrace{\vec u\cdot(\vec u\times\vec v)}_{=0}=\delta\|u\|^2$$so $\vec u\cdot\vec w=\alpha$ only for $\delta=\frac{\alpha}{\|u\|^2}$. Therefore if $\vec u,\vec v$ are perpendicular, non zero vectors, a solution to $\vec u\times\vec w=\vec v$ and $\vec u\cdot\vec w=\alpha$ exists and is $$\vec w=\frac{\alpha}{\|u\|^2}\vec u-\frac{1}{\|u\|^2}(\vec u\times\vec v)$$

Edit Regarding the notation: $\|u\|^2$ is $u\cdot u$, so not exactly the length of the vector, but its square power.

Regarding $\vec u\times\vec w$. We said that $\vec w$ is in the form $\frac{\alpha}{\|u\|^2}\vec u-\frac{1}{\|u\|^2}(\vec u\times\vec v)$, so $$\vec u\times\vec w=\vec u\times\left(\frac{\alpha}{\|u\|^2}\vec u-\frac{1}{\|u\|^2}(\vec u\times\vec v)\right)=\frac{\alpha}{\|u\|^2}\underbrace{\vec u\times\vec u}_{=0}-\frac{1}{\|u\|^2}\vec u\times(\vec u\times\vec v)$$ so we need to prove that $$-\frac{1}{\|u\|^2}\vec u\times(\vec u\times\vec v)=\vec v$$ Now, if $\vec u=(u_1,u_2,u_3)$ and $\vec v=(v_1,v_2,v_3)$, then $\vec u\times\vec v$ is defined in coordinates as follows $$\vec u\times\vec v=\begin{pmatrix} u_2v_3-u_3v_2\\ u_3v_1-u_1v_3\\ u_1v_2-u_2v_1 \end{pmatrix}$$ and $$\vec u\times(\vec u\times\vec v)=\begin{pmatrix} u_2(u_1v_2-u_2v_1)-u_3(u_3v_1-u_1v_3)\\ u_3(u_2v_3-u_3v_2)-u_1(u_1v_2-u_2v_1)\\ u_1(u_3v_1-u_1v_3)-u_2(u_2v_3-u_3v_2) \end{pmatrix}$$ here you simplify the coordinates for $\vec u\times(\vec u\times\vec v)$ using the hypotesis that $\vec u,\vec v$ are perpendicular, that is, $u_1v_1+u_2v_2+u_3v_3=0$. They are all done in the same way, so consider just the first coordinate $u_2(u_1v_2-u_2v_1)-u_3(u_3v_1-u_1v_3)$: $$u_2(u_1v_2-u_2v_1)-u_3(u_3v_1-u_1v_3)=u_2u_1v_2-u_2^2v_1-u_3^2v_1+u_3u_1v_3=-(u_1^2+u_3^2)v_1+u_1(u_2v_2+u_3v_3)=\star$$now you use $\vec u\perp \vec v$, in particular $u_2v_2+u_3v_3=-u_1v_1$, so $$\star=-(u_1^2+u_3^2)v_1-u_1^2v_1=-\|u\|^2v_1$$In the same way you simply all the other coordinates of $\vec u\times(\vec u\times\vec v)$ and you get, as expected $$\vec u\times(\vec u\times \vec v)=-\|u\|^2\vec v\rightarrow-\frac{1}{\|u\|^2}\vec u\times(\vec u\times\vec v)=\vec v$$ which is what we wanted.