Given $a,b,k,l$, solve $k\vec{xa}+l\vec{xb}=m\vec{xc}$ for $m$ and $c$

Question

Given two points $a$ and $b$ in a Euclidean space, and two nonzero real numbers $k,l\in\Bbb R-\{0\}$, how to find a point $c$ in this space and a nonzero real number $m\in\Bbb R-\{0\}$ such that $k\vec{xa}+l\vec{xb}=m\vec{xc}$ for all $x$ in the space? Proof of existence of the pair $c$ and $m$ is also ok if a closed-form particular solution is impossible.

I just drew a picture and got stuck by this question, because I could not see a fixed point $c$ that the vector combination $k\vec{xa}+l\vec{xb}$ always points to. Thank you for any clue for me to proceed.

Answer 1

Let $a=\left( a_1,a_2,\cdots ,a_n \right) ,b=\left( b_1,b_2,\cdots ,b_n \right) ,c=\left( c_1,c_2,\cdots ,c_n \right), x=\left( x_1,x_2,\cdots ,x_n \right) $. Then $k\vec{xa}+l\vec{xb}=m\vec{xc}$ is equivalent to say $$k\left( a_1-x_1,a_2-x_2,\cdots ,a_n-x_n \right) +l\left( b_1-x_1,b_2-x_2,\cdots ,b_n-x_n \right) =m\left( c_1-x_1,c_2-x_2,\cdots ,c_n-x_n \right) $$

EDIT: I was wrong. The correct way to proceed is as the comment shows. $m$ should be $k+l$, in order that $$[m-(k+l)]x_i=mc_i-(ka_i+lb_i)$$ holds for all $x$.

Answer 2

The equations are linear, so if it is true for $x=(1,0)$ and $x=(0,1)$ it is true everywhere. If you write it componentwise you have two equations in two unknowns, which you should be able to solve most of the time. Do your conditions prohibit the singular case?