Given the following vector equation in three dimensions
$\mathbf{r} + (\mathbf{r} \times \mathbf{d}) = \mathbf{c}$
where $\mathbf{c}$ and $\mathbf{d}$ are fixed given vectors, how can you find all solutions for $\mathbf{r}$?
So far I have tried the following steps, to show that $\mathbf{r}$ must lie in a given plane.
By taking the dot product with $\mathbf d$ on both sides, we obtain
$$\mathbf{r} + (\mathbf{r} \times \mathbf{d}) = \mathbf{c} \implies (\mathbf{r} + (\mathbf{r} \times \mathbf{d})) \cdot \mathbf{d} = \mathbf c \cdot \mathbf d.$$
Since $\mathbf r \times \mathbf d$ is perpendicular to $\mathbf d$, their dot product is zero, so we get the following (Equation $\ast$)
$$\mathbf r \cdot \mathbf d = \mathbf c \cdot \mathbf d$$
from which we can deduce that $\mathbf r$ (as a position vector) lies in the plane that contains $\mathbf c$ and is normal to the vector $\mathbf d$.
However this doesn't necessarily imply that all points in this plane are valid solutions for $\mathbf r$. I can't see how Equation ($\ast$) can be substituted back into the original equation to somehow eliminate a term in $\mathbf r$ or simplify it. How do you solve this equation, making sure that you find all solutions for $\mathbf r$?
Best Answer
You have 3 direction in space. Assuming $\mathbf d$ and $\mathbf c$ are not collinear (in this case $\mathbf r$ would be in the same direction), then we can use the three directions to be $\mathbf c$, $\mathbf d$, and $\mathbf c\times\mathbf d$. The first two might not be perpendicular. Then write $$\mathbf r=\alpha\mathbf c+\beta\mathbf d+\gamma\mathbf c\times\mathbf d$$ Plug this into your equation, then multiply the equation by each of the vectors of the basis.