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.
 A: 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 $$r=\dfrac{c+(c\cdot d)\, d-c\times d}{1+\lVert d\lVert^2}$$
A: In terms of vectors and matrices, the equation can be rewritten as
$$
(I-[\mathbf d]_\times)\mathbf r=\mathbf c.
$$
By a change of orthonormal basis, we may assume that $\mathbf d=(0,0,a)^T$. Therefore
$$
[\mathbf d]_\times=\pmatrix{0&-a&0\\ a&0&0\\ 0&0&0}
$$
and
$$
(I-[\mathbf d]_\times)^{-1}
=\pmatrix{1&a&0\\ -a&1&0\\ 0&0&1}^{-1}
=\pmatrix{\frac{1}{1+a^2}&\frac{-a}{1+a^2}&0\\ \frac{a}{1+a^2}&\frac{1}{1+a^2}&0\\ 0&0&1}
=\frac{1}{1+\|\mathbf d\|^2}\left(I+\mathbf d\mathbf d^T+[\mathbf d]_\times\right).
$$
It follows that
$$
\mathbf r
=(I-[\mathbf d]_\times)^{-1}\mathbf c
=\frac{1}{1+\|\mathbf d\|^2}\left(\mathbf c+(\mathbf d\cdot \mathbf c)\mathbf d+\mathbf d\times\mathbf c\right).
$$
A: Another approach, which is equivalent to already presented answers, is to use Levi-Civita symbol to express vector $r$ explicitly. The following equation
$$
\vec r + \vec r \times \vec d = \vec c
$$
can be written as
$$
r_i + \varepsilon^{ijk} r_j d_k = c_i,
$$
which leads to the system of equations
$$
\begin{cases}
r_1 + r_2 d_3 - r_3 d_2 = c_1\\
r_2 + r_3 d_1 - r_1 d_3 = c_2\\
r_3 + r_1 d_2 - r_2 d_1 = c_3
\end{cases}.
$$
Finally, we get three equations for three variables $r_1, r_2, r_3$, which can be solved using the standard methods.
