How to make 4 vectors (all which you know the direction but you can set the magnitude) add up to a known vector? So if you have 4 vectors, and you know their directions (which all could be different), and you could change all their magnitudes, what would you set the magnitudes equal to if you want all 4 vectors to add up to another vector which you know?
This might seem like a little bit too specific question, but the reason I ask this is for a program I'm making with kRPC for KSP in which 4 tilted engines hover an aircraft, even when the entire aircraft is tilted. I tried searching it, but I didn't know exactly what to search. I don't know a lot about the math of vectors. Thanks!
 A: I'll assume that we're working in $3$-D space. Call the four given vectors $\vec v_1, \vec v_2, \vec v_3, \vec v_4 \in \mathbb R^3$, and let $\vec b \in \mathbb R^3$ be the fifth vector that we want to express as a linear combination of the first four. Then we seek weights $c_1, c_2, c_3, c_4 \in \mathbb R$ such that:
$$
c_1\vec v_1 + c_2 \vec v_2 + c_3 \vec v_3 + c_4 \vec v_4 = \vec b
$$
In fancier linear algebra terms, we want to know if $\vec b \in \text{Span}\{\vec v_1, \vec v_2, \vec v_3, \vec v_4\}$. Deciding whether this is true or not amounts to solving a linear system of three equations in four unknowns (which can be accomplished by forming an augmented matrix and solving via Gaussian elimination). It turns out that if at least three of the four vectors are linearly independent (that is, some three vectors form a basis for $\mathbb R^3$), then we are guaranteed a unique solution for the weights (where the fourth unused vector is redundant and assumed to be given a weight of zero). If a solution exists and we want to use all four vectors, then note that there will be infinitely many combinations of weights that can be used to form the desired linear combination.
