The problem asks: Find parametric equations and an equation in vector form for the line in $R^{2}$ with the equation: $y = 3x -1$
vector form is $x = p + td$ where $p$ and $d$ are vectors and $t$ is a scalar. To find $p$ I pick a number for $x$ and plug into the equation to get the corresponding $y$. So $x = 0$ gives $y = -1$, so $p = [0, -1]$ because 3 is the scalar multiplied with $x$ and 1 is the scalar multiplied with $y$, I thought the vector form would be:
$$
\begin{bmatrix} x\\ y\\ \end{bmatrix} = \begin{bmatrix}
0 \\
-1 \\
\end{bmatrix} + t
\begin{bmatrix}
3 \\
1 \\
\end{bmatrix}
$$
and from the vector form the parametric form follows easily, so I'm going to skip writing it. This however is wrong since the answer sheet gives the correct answer with $3$ going to y and 1 going to $x$:
$$
\begin{bmatrix} x\\ y\\ \end{bmatrix} = \begin{bmatrix}
0 \\
-1 \\
\end{bmatrix} + t
\begin{bmatrix}
1 \\
3 \\
\end{bmatrix}
$$
I don't understand why that's so, could someone please explain. Also, where does the -1 go in $y = 3x -1$, how come it isn't represented in any of those forms.
Best Answer
At least for an intuitive approach as to why the answer sheet's answer is right, note that the line $y = 3x - 1$ has slope 3. Your equation parametrizes a line that has slope 1/3, since if you let $t$ vary from 0 to 1, say, $x$ varies from 0 to 3, while $y$ varies only from -1 to 0; this is a line of slop 1/3, not of slope 3.
The answer key equation gives a line that, as $t$ varies from 0 to 1 (say), has $x$ vary from 0 to 1 and $y$ vary from -1 to 2, which is the correct slope of 3.
The directional vector $d$ should be the slope vector; if the slope, for instance is 3 (as it is in this case), any vector that has slope 3 will do. So just pick the vector $$ d = \begin{pmatrix} 1 \\ 3 \end{pmatrix}. $$
As for the -1, note that at $x = 0$, we have $y = 3\cdot 0 - 1 = -1$. This is captured in the vector form of the equation by the initial position vector $$ p = \begin{pmatrix} 0 \\ -1 \end{pmatrix}. $$ Because look what happens when we put $t = 0$ in the vector equation; we get $(x, y) = (0, -1) + 0\cdot d = (0, -1)$, meaning that the point $(0, -1)$ is on our line, as it should be. Really, it just serves as a convenient starting point to draw the vector equation line from.