We refer to the same picture as in Exercise 1. This time the coordinates of the points A(4, 0, 0), B(4, 6, 0), C(0, 6, 0) and M(2, 3, 6) are given.
b) Find the parametric equation of the line AF.
I thought I solved it correctly but somehow I missed a step? This is my solution so far. From the coordinates I got F(2, 6, 6) So I got the par. equation like this: $\vec{r}=\vec{OA}+t.\vec{AF}$
$\vec{AF}= \begin{pmatrix}-2\\ 6\\ 6\end{pmatrix}$ so $\vec{r}=$ $\begin{pmatrix}4\\ 0\\ 0\end{pmatrix}+t\cdot \begin{pmatrix}-2\\ 6\\ 6\end{pmatrix}$
But my solution is wrong, it's $\vec{r}=$ $\begin{pmatrix}4\\ 0\\ 0\end{pmatrix}+t\cdot \begin{pmatrix}-1\\ 3\\ 3\end{pmatrix}$ instead, why??
Your answer and the solution are basically the same, with the solution using a second vector that is $\frac{1}{2}$ of yours, plus the range of the set of resulting values are the same.
You can see this equivalence by replacing their $t$ with $2t'$, moving the factor of $2$ into the vector $\begin{pmatrix}-1\\ 3\\ 3\end{pmatrix}$ so it becomes your vector $\begin{pmatrix}-2\\ 6\\ 6\end{pmatrix}$, and then replace your $t$ with $t'$. In both cases, you will then get $\vec{r}=$ $\begin{pmatrix}4\\ 0\\ 0\end{pmatrix}+t'\cdot \begin{pmatrix}-2\\ 6\\ 6\end{pmatrix}$
As for why the solution removed the factor of $2$, I don't know for sure, but perhaps it was a desire to remove the $\gcd(-2,6,6) = 2$ common factor among the vector elements.