Suppose we have a vector $a$ in $\mathbb R^3$ and an unknown vector $v$, but we know $a \cdot v$ and $a \times v$. Can we find $v$? How?
Sources: Based on Shifrin's Multivariable Mathematics and MIT OCW 18.013A.
My solution is below, to which I request verification and suggestions. Note: Other solutions exist, such as here; this question is to verify and critique my solution.
Let $$a,v,c \in \mathbb R^3 \\ d \in \mathbb R \\ a \cdot v = d \\ a \times v = c.$$ Given $a,c,d$, find $v$.
Any $v$ can be decomposed into $v = v_{\parallel} + v_{\perp}$, where $v_{\parallel} \parallel a, v_{\perp} \perp a$. Therefore
$$\begin{align} v_{\parallel} &= \frac d {a \cdot a} \, a \\ \|v_{\perp}\| &= \frac {\|c\|}{\|a\|}\\ \end{align}$$ since parallelogram area $\|c\|$ is base $\|a\|$ times height $\|v_{\perp}\|$.
Since $c$ is perpindicular to both $a$ and $v$, we have $$\begin{align} c \cdot v &= 0 \\ c \cdot (v_{\parallel} + v_{\perp}) &= 0 \\ c \cdot v_{\perp} &= 0 \end{align}$$ and $v_{\perp}$ is $\perp$ to both $a$ and $c$ and therefore parallel to $a \times c$:
$$\widehat{v_{\perp}} = \pm \frac {c \times a}{\|c \times a\|}.$$
In fact, $\widehat{v_{\perp}} = \frac {c \times a}{\|c \times a\|}$. To see this, consider the case where $a = i$ and $v$ is in the first or second quadrant. Then $\hat c = k$ and $\widehat{v_{\perp}} = j = \frac {c \times a} {\| c \times a\|}$. Since cross product is invariant under rotation, any $a, c$ and $v$ are equivalent to this.
This gives us $$v = \frac d {a \cdot a} \, a + \frac {\|c\|}{\|a\| \|c \times a\|} \, c \times a.$$
Update: which, as Ted Shifrin and ghasheem point out, simplifies to
$$\begin{align}v &= \frac d {a \cdot a} \, a + \frac {c \times a}{\|a\|^2}\\ v &= \frac {da + c \times a} {a \cdot a}.\end{align}$$
Since $ a \times v = c$, then $c$ must be orthogonal to $a$, otherwise there is no solution. Vector $v$ is orthogonal to $c$, so it lies in the plane spanned by $ a $ and $ a \times c $
$ v = \alpha a + \beta (a \times c) $
Substituting $v$ into $ a \times v = c $ gives
$ \beta ( a \times (a \times c) ) = c $
Now,
$ a \times (a \times c) = a ( a \cdot c) - c (a \cdot a ) $
The first term is zero because $c$ is orthogonal to $a$. Therefore
$ \beta = - \dfrac{1}{ (a \cdot a ) } $
while $\alpha$ is determined from the dot product with $a$
$ v \cdot a = \alpha (a \cdot a) = d $
so $ \alpha = \dfrac{d}{a \cdot a } $
Therefore,
$ v = \dfrac{1}{a \cdot a } ( d a - (a \times c) ) $