Calculate $\vec{M}$ if $\vec{M} \cdot \vec{A} = c$

Question

I am trying to solve an equation that involve the inner product of 2 vectors and I was wondering if there was a way to solve it.

I know that $$ \vec{M} \cdot \vec{A} = c $$ Where $c$ is a constant scalar and vectors are in 2D space, Then how do I compute $\vec{M}$?

Answer 1

Since these are 2d vectors, we may write $M=(M_{1},M_{2})$, $A=(A_{1},A_{2})$, so your equation is $$M_{1}A_{1}+M_{2}A_{2}=c \implies M_{1}=\frac{c-M_{2}A_{2}}{A_{1}}$$ Therefore, if we let $t$ be any real number, the vector $$M=\pmatrix{\frac{c-tA_{2}}{A_{1}}\\t}=\pmatrix{\frac{c}{A_{1}}\\0}+t\pmatrix{-A_{2}/A_{1}\\1}$$ satisfies your equation. There are many choices (as you should expect! If $c=0$, there are many vectors perpendicular to, say, $(1,0)$ - just take scalar multiples of them).

Answer 2

In $\mathbb{R}^3$, if $\vec M=(x,y,z)^T$ and $\vec A=(a_1,a_2,a_3)^T$, than $\vec M \cdot \vec A=c$ is the equation of a plane: $$ a_1x+a_2y+a_3z=c $$ that is the plane orthogonal to $\vec A$ and whose distance from the origin is $d=\frac{|c|}{\sqrt{a_1^2+a_2^2+a_3^2}}$.

So, given $\vec A$ and $c$ , $\vec M$ can be any vector in this plane.