Vectors basic concepts Given vectors ${p}$ and ${d}$, we can describe the line through ${p}$ in direction ${d}$ as the vectors $x$ that satisfy
${x} = p + t d.$
In this problem we explore another representation for lines.
a) Let $p = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and ${d} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$. Find vectors $q$ and $n$ such that the line described above can also be described as the vectors $x$ that satisfy
$(x - q) \bullet n = 0.$
b) Show that for any ${p}$ and nonzero ${d}$, there exist vectors ${q}$ and ${n}$ so that a vector ${x}$ lies on the line
${x} = {p} + t{d}$
if and only if
$({x} - {q}) \bullet {n} = 0.$
Hi, I have tried expressing all the vectors with variables, but have gotten nowhere.  Any help is appreciated!
 A: For simple case in $\mathbb{R}^2$
Let $x=\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right)$, $p=\left(\begin{matrix} p_1 \\ p_2 \end{matrix}\right)$, $d=\left(\begin{matrix} d_1 \\ d_2 \end{matrix}\right)\ne \bar0$
Then equation of line is given by $\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right)=\left(\begin{matrix} p_1 \\ p_2 \end{matrix}\right)+t\left(\begin{matrix} d_1 \\ d_2 \end{matrix}\right)$
$\iff\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right)=\left(\begin{matrix} p_1 + td_1 \\ p_2 + td_2 \end{matrix}\right)$
Solving for t, we get $t=\frac{x_1-p_1}{d_1}=\frac{x_2-p_2}{d_2}$
$\iff\frac{x_1-p_1}{d_1}-\frac{x_2-p_2}{d_2}=0 \iff (x_1-p_1)d_2-(x_2-p_2)d_1=0$ 
$\iff \left[\left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right)-\left(\begin{matrix} p_1 \\ p_2 \end{matrix}\right)\right].\left(\begin{matrix} d_2 \\ -d_1 \end{matrix}\right)=\bar0$
$\iff (x-q).n=\bar0$, where $q=\left(\begin{matrix} p_1 \\ p_2 \end{matrix}\right), n=\left(\begin{matrix} d_2 \\ -d_1 \end{matrix}\right)$
You can solve (a) following the same steps.
