Solve the equation: $a \times c = b + c$ for linearly independent vectors a and b.

Question

$a$, $b$ and $c$ are all vectors. $a \times c$ is the cross product of $a$ and $c$. After making a quick sketch, I found that $b$, $c$ and $a \times c$ form a right triangle, which I was not sure how to use to come to a solution. I ended up getting that $a=c$ and $c=-b$, but I'm pretty sure that is wrong. Another question is whether the answer also holds true for an arbitrary pair of vectors.

Answer 1

Since $a$ and $b$ are linearly independent, $a\times b$ is non-zero and not a multiple of either. Therefore $\{a, b, a\times b\}$ forms a basis for three dimensional space. So any $c$ must have a unique expression as $$c = xa + yb + z(a\times b)$$ for some scalars $x,y,z$. Now $$a\times c = x(a\times a) + y(a\times b) + z(a\times(a\times b)) = z(a\cdot b)a - z\|a\|^2b+y(a \times b)$$ since $a\times a = 0$ and making use of the vector triple product rule. Also, $$b + c = xa + (y+1)b + z(a\times b)$$ Equating these two, and again making use of the uniqueness of representation in a basis, we get $$x = z(a\cdot b)\\y + 1 = -z\|a\|^2\\z = y$$ Which solves to $$c = \frac {-1}{1+\|a\|^2}\left((a\cdot b)a + b + a\times b\right)$$

When $a$ and $b$ are not linearly dependent, then $a = mb$ for some scalar $m$. See if you can figure what is true in this case.