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I received a question about a girl who can ride a hover board and a magic carpet whose restricted to move in the "movement vector" [3 1] and [1 2] respectively. The "movement" vectors represent how much the vehicle moved "forward" in one hour. For instance, the hover board moves 3 units east and 1 unit north in that hour.

Determine whether the girl can travel anywhere about the 2D plane assuming that she starts from the origin.

My answer to this is that the girl cannot move around the entire span of the 2D plane. I noticed that the vectors given are just velocity vectors of the vehicle so the equation that describes the location she can be at is $$\begin{bmatrix} 3 \\ 1 \\ \end{bmatrix} t_1+\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}t_2=\begin{bmatrix} x \\ y \\ \end{bmatrix}$$ This obviously requires a small understanding of velocity but we assume this is true for now.

At first I thought I can just prove linear independence and say that the span of these set of vectors to be 2D space but $ t_1 >= 0, t_2>=0$ since we can't travel back in time.

My question is how do I prove algebraically the span of these two vectors with the constraint $ t_1 >= 0, t_2>=0$?

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  • $\begingroup$ Surely you cannot get $(x,y)=(-1,-1)$ under the constraint $t_1$, $t_2\ge0$. $\endgroup$ – Lord Shark the Unknown Feb 10 at 15:23
  • $\begingroup$ Yes, but how do we prove the span algebraically under the constraints? $\endgroup$ – Nick Yarn Feb 10 at 15:30
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For instance she can't get to $(1,0)$, since $t_1(3,1)+t_2(1,2)=(1,0)\implies t_2=-\frac15$.

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Since $t_1\ge0$ and $t_2\ge0$ and since we can solve to get

\begin{eqnarray} t_1&=&\,\,\,\frac{2}{5}x-\frac{1}{5}y\ge0\\ t_2&=&-\frac{1}{5}x+\frac{3}{5}y\ge0 \end{eqnarray}

it follows that

$$ \frac{x}{3}\le y\le2x $$

Span

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