solving vectors geometrically

[centenial]

I'm sorry if this is in the wrong forum. A bunch of questions have come up like this for my linear algebra course. I know how to solve matrices using Gauss-Jordan elimination, but I'm not positive about solving this geometrically.

I know that when you add vectors geometrically, you can line v1 and v2 up so that v3 is the diagonal. So, my intuition would tell me that you could have infinitely many solutions because of variations in the scalers x and y. Is this the right direction?

Any help would be much appreciated! Thanks!

 

[soroban]

Helolo, centenial!

\(\displaystyle \text{Consider the vectors }\vec v_1,\,\vec v_2,\,\vec v_3\text{ shown in the accompanying sketch.}\)

\(\displaystyle \text{How many solutions has the system: }\:x\!\cdot\!\vec v_1 + y\!\cdot\!\vec v_2 \:=\:\vec v_3\)
\(\displaystyle \text{Argue geometrically.}\)

          D . . . . . . . . . * C
           .              *  .
          .           *     .
         .        * v3     .
        *     *           .
    v2 *  *              .
      *  *  *  *  . . . . B
    A     v1

\(\displaystyle \text{Since }\vec v_1\text{ and }\vec v_2\text{ are given, there is }one\text{ parallelogram that produces the resultant }\vec v_3.\)

\(\displaystyle \text{We want scalars }x\text{ and }y\text{ such that: }\:\begin{Bmatrix}x\!\cdot\!\vec v_1 \;=\; \overrightarrow{AB} & \Rightarrow & x \;=\; \dfrac{\left|\overrightarrow{AB}\right|}{|\vec v_1|} \\ \\[-2mm] y\!\cdot\!\vec v_2 \;=\; \overrightarrow{AD} & \Rightarrow & y \;=\; \dfrac{\left|\overrightarrow{AD}\right|}{|\vec v_2|} \end{Bmatrix}\)

\(\displaystyle \text{Therefore, there is }one\text{ solution.}\)