Finding set of vectors satisfying $\textbf r+\textbf r\times\textbf d=\textbf c$ Let $\mathbf{c}$ and $\mathbf{d}$ be fixed vectors in $\mathbb{R^3}$. Find all vectors $\textbf r$ such that $\textbf r+\textbf r\times\textbf d=\textbf c$.
Attempt: Take the vector product with $\textbf r$: $\textbf r\times\textbf r+\textbf r\times(\textbf r\times\textbf d)=\textbf r\times\textbf c$
$\Rightarrow \textbf r\times\textbf r+(\textbf r\cdot\textbf d)\textbf r-(\textbf r\cdot\textbf r)\textbf d=\textbf r\times\textbf c$
$\Rightarrow\textbf r\times\textbf r=\textbf r\times\textbf c$
$\Rightarrow \textbf r\times\textbf c=0$.
Hence any vector parallel to the vector $\textbf c$ will work.

You have $r\times d$ here, so to solve for $r$ you need to calculate the dot product and the cross product with $d$ instead of $r$.

*

*$(r \cdot d)+(r\times d) \cdot d=(r \cdot d)+0=(r \cdot d)=(c \cdot d)$


*$r\times d+(r\times d)\times d=(c-r)+(r \cdot d)\,d−(d \cdot d)\ r=c\times d$
Which allows to isolate $r$.
$(1+\lVert d\lVert^2)\,r=c-c\times d+(r\cdot d)\,d=c-c\times d+(c\cdot d)\, d$
Therefore $$r=\dfrac{c+(c\cdot d)\, d-c\times d}{1+\lVert d\lVert^2}$$
