0
$\begingroup$

Say I have two vectors, $A$ and $B$. They have magnitude $|A|$ and $|B|$, and angle $\alpha$ between them (where $A\cdot B=(AxBx + AyBy)$ and $\cos(alpha) = \frac{A\cdot B}{|A||B|}$.

Now, say I have another vector, $A'$ and I know there exists another vector $B'$, such that $A$ and $B$ are scaled and/or rotated by the same matrix, $M$. So, $A' = MA$ and $B' = MB$.

I know each pair share the same angle (I.e., $\alpha(A,B)=\alpha(A',B')$) and the proportion of the magnitudes must be the same ($\frac{|A|}{|C|} = \frac{|B|}{|D|}$).

However, the plot thickens! $A'$ can have four possible vectors that satisfy the conditions for $B'$ (as shown by the dotted lines in the sketch). None of these are invalid representations.

I have tried using and rearranging the above equations in many many ways, to absolutely no avail. Representing points as matrices , $$A' = MA \Rightarrow M=A(A'^{-1}) \Rightarrow B'=A(A'^{-1})B $$ seemed of interest, but I am really lost!

So, is there an equation (I would guess its a linear map?) to get the coordinates for all vectors that satisfy the conditions of $B'$? I think it may involve plus/minus the distance from the center of the vector, to get all four equations, as that visibly is where there is symmetry of all four possibilities.

I'm very new to linear algebra (determinants and inverse matrices are the limit of my knowledge), so please try to make your answer slightly layman if it involves anything complex! (no imaginary number pun intended there, either :p)

line transformations

  • 0
    By "lines" you probably mean "vectors"? Lines generally don't have magnitudes... – 2017-02-14
  • 0
    Ah yes I do! They are lines in the sense of A = [ P(x,y), Q(x,y)] where P and Q are the two points of the line. Which is more like a vector off of the origin! – 2017-02-14
  • 0
    What do you mean by $A'^{-1}$? $A'$ is a vector. – 2017-02-14
  • 0
    Representing the vectors as transformations, sorry. I can define that quickly in an edit! – 2017-02-14
  • 0
    The “vectors” that you’ve drawn look like undirected line segments to me. Vectors have a direction, and taking that direction into account with immediately eliminate two of the possibilities for $B'$. In addition, it looks from your diagram that there are translations involved as well, in which case there will be an infinite number of possibilities for $B'$ (and you need to look at *affine* transformations). You might want to sharpen your question to include all of the conditions/constraints explicitly. – 2017-02-14
  • 0
    Okay, again please note I'm very new to linear algebra (and general math!) as I stated at the bottom. I also don't understand how, if A is translated to B, how there can be infinite transformations from A' to B', but there is certainly only four lines I wish to accept. Does this help narrow down the question enough to be answered? – 2017-02-14

0 Answers 0