I need to solve $k_b$, $k_a$ and $t$ in this equation in a program I'm creating: $$k_b \bar b - k_a \bar a + \frac{1}{t} \cdot k_a \bar a - \frac{1}{t}(\bar c - \bar a_1) = \bar a_1 - \bar b_1$$ All the vectors are known (they are 3D vectors). I was planning to make a system of equations where I'd split this equation in three by the x, y and z components of the vectors, and then use Gaussian elimination to solve it. But as I understand it, that would need each term on the left side to be of the form $variable \cdot coefficient$, so that I could make a matrix of the coefficients. But I don't know what to do with the third term where $\frac{1}{t}$ is multiplied by $k_a$.
Is there some way to get this equation to a simpler form? Is it possible to use Gaussian elimination for a system of equations like this? Is there another algorithmic way to solve these equations?
Here is a way to calculate $t$ when $\vec a$ is not parallel to $\vec b$
Define $\vec d \equiv \vec c - \vec a_1$ and $\vec e \equiv \vec a_1 - \vec b_1$ and $\vec f\equiv \vec a \times \vec b$
multiply through by $t$ to get .. $$tk_b \vec b + (1-t)k_a \vec a = \vec d + t\vec e \tag{1}$$ taking the dot product with $\vec f$ we have ... $$ 0 = \vec d \cdot \vec f+t(\vec e \cdot \vec f)$$ So that $$ t =- \frac{\vec d \cdot \vec f}{\vec e \cdot \vec f}$$ note that no solution exists if $(\vec d \cdot \vec f)=0$ or $(\vec e \cdot \vec f) =0$
If $\vec f=0$ then solutions only exist if $(\vec d \times \vec b)\times (\vec e \times \vec b)=0$ and $(\vec d \times \vec a)\times (\vec e \times \vec a)=0$.