The funny thing about these approaches to lines is that they have an extra variable, the parameter $t$, and are not simply a relationship equation between $x$ and $y$. Of course, the more familiar equation for a line in the plane is such a relationship equation: the point $(x,y)$ is on the line in question if $$y=mx+b$$ for some constants $m$ and $b$. Sometimes we write this as $$Ax+By=C$$ where $A=m$, $B=-1$, and $C=-b$.
We can turn the parametric equations into a relationship equation like that one by solving both equations for $t$: $$t=\frac{1}{v_1}\left(x-d_1\right)$$ and $$t=\frac{1}{v_2}\left(y-d_2\right)$$ and setting them equal $$\frac{1}{v_1}\left(x-d_1\right)=\frac{1}{v_2}\left(y-d_2\right).$$ Again, we can write this as $$Ax+By=C$$ where $A=\frac{1}{v_1}$, $B=-\frac{1}{v_2}$, and $C=d_1-d_2$ (check the algebra!)
Now let's transform this a little, to be a relationship equation which will tell us when vector $\vec{r}=\left<x,y\right>$ is on our line: $$Ax+By=\vec{n}\cdot\vec{r}=C,$$ where the vector $\vec{n}=\left<A,B\right>$. That's interesting! If the $C$ had been $0$, I would recognize this equation $$\vec{n}\cdot\vec{r}=0$$ as saying that $\vec{r}$ is perpendicular (oh, wait, let's be fancy: orthogonal) to $\vec{n}$. And, if you think about it, the set of all vectors orthogonal to a fixed vector $\vec{n}$ is a line through the origin!
The only things we have to do are get equations for lines which don't happen to go through the origin, and get back that term $C$, when it is not $0$. — These happen to be the same thing, hooray! — The idea is simply to take a displacement vector $\vec{d}$ from the origin to any point on the line. Then any vector to a point on the line $\vec{r}$ will, if we move it over by looking at $\vec{r}-\vec{d}$, be a vector to a point on the parallel line through the origin. That is, $\vec{r}-\vec{d}$ must be a solution to the above equation with a $0$ on the right side. In other words, $$\vec{n}\cdot\left(\vec{r}-\vec{d}\right)=0$$ or, equivalently, $$\vec{n}\cdot\vec{r}=\vec{n}\cdot\vec{d},$$ which is the normal form of the equation for the line in the plane with normal vector $\vec{n}$ and displacement vector $\vec{d}$. Sometimes we write this out by naming the components of these vectors as \begin{align} \vec{n}&=\left<a,b\right>,\\ \vec{r}&=\left<x,y\right>, \text{ and}\\ \vec{d}&=\left<d_1,d_2\right> \end{align} so that $$ ax+by=(ad_1+bd_2) $$ is another equation which we also call the normal form of the equation of a line.
Example 1: Suppose we want to
Find the normal equation of the line in the plane which passes through $(1,1)$ and is parallel to the line $y=2x+1$.To write down the normal equation of the line, we need a normal vector $\vec{n}$ and a point $\vec{d}$ on the line.
We were given the $\vec{d}=\left<1,1\right>$, so we only need to figure out the $\vec{n}$.
Now parallel lines go in the same direction, so they have the same normal vectors. That means if we can figure out a normal to the given line $y=2x+1$, it will do for the normal we're looking for.
That given line has slope $2$, so a vector in that direction would be $\vec{v}=\left<1,2\right>$ (the $y$ component is the rise, the $x$ component is the run, in the definition "slope is rise over run".
Hmm, some random vector orthogonal to this $\vec{v}$ will do, such as $\vec{n}=\left<2,-1\right>$.
This gives $$\vec{n}\cdot\vec{r}=\vec{n}\cdot\vec{d}$$ as the normal form of the equation of the desired line. Actually, that right hand side is just the number $\vec{n}\cdot\vec{d}=\left<2,-1\right>\cdot\left<1,1\right>=1$, so the normal form of the equation of the desired line is either $$\vec{n}\cdot\vec{r}=1$$ or, using components $\vec{r}=\left<x,y\right>$, $$2x-y=1 .$$
Now we want a normal form for this.
Suppose we take the normal form for lines and just write it down with vectors in space, having three components each? That would be $$\vec{n}\cdot\vec{r}=\vec{n}\cdot\vec{d},$$ and, as usual, if $\vec{d}=\vec{0}$, the plane would go through the origin.
And this makes perfect sense, since $\vec{n}\cdot\vec{r}=0$ is simply the equation satisfied by all vectors $\vec{r}$ orthogonal to $\vec{n}$, and in three-dimensional space, the set of vectors orthogonal to one fixed vector forms a plane! Hence we are left with the normal form of the equation for the plane in space with normal vector $\vec{n}$ and displacement vector $\vec{d}$ being $$\vec{n}\cdot\vec{r}=\vec{n}\cdot\vec{d}$$ or, if we write this out by naming the components of these vectors as \begin{align} \vec{n}&=\left<a,b,c\right>,\\ \vec{r}&=\left<x,y,z\right>, \text{ and}\\ \vec{d}&=\left<d_1,d_2,d_3\right> \end{align} we also refer to $$ ax+by+cz=(ad_1+bd_2+cd_3) $$ as the normal form of the equation of a plane.
In practice, this form is convenient to write down, it is easy to use to find points on the plane — just plug in some numbers for two of the three variables, solve for the last one and you'll have a point on the plane — and to find equations of planes parallel to given ones — parallel planes have the same normals — but it can be hard to find the normal vector $\vec{n}$! At this point in the course, the only cases in which we can do this are where we can see the geometry well enough to write down the normal vector pretty much by inspection. Later, we will get a tool (the cross product) which will make this easier....
Example 2: Suppose we want to
Find the normal equation of a plane which slices the unit cube nestled in the corner of the first octant down from the edge connecting $(0,0,1)$ to $(1,0,1)$ down to the opposite edge, which connects $(0,1,0)$ to $(1,1,0)$.Well, we can simply read off the displacement vector as going from the origin to any of those four given points, such as $\vec{d}=\left<0,0,1\right>$, for example.
Notice that the two given line segments (which lie in the plane), are parallel to the $x$-axis. That means the normal vector to the plan should be orthogonal to the $x$-axis, so it will have an $x$-component of $0$. If you can sketch or visualize this plane clearly enough, you will see that it's intersection with the $yz$-plane is the line $z=1-y$. Therefore a normal vector could be $\vec{n}=\left<0,1,1\right>$.
Hence the desired normal form of $\vec{n}\cdot\vec{r}=\vec{n}\cdot\vec{d}=\left<0,1,1\right>\cdot\left<0,0,1\right>=1$ is $$y+z=1 .$$
Example 3: How about
Given a plane whose normal form equation is $x+2y+3z=2$, find a vector and parametric forms of the equation of the plane.We're going to need a displacement vector $\vec{d}$ from the origin to the plane and two vectors $\vec{u}$ and $\vec{v}$ in the plane.
If we put $y=z=0$, the normal form equation simplifies to $x=2$. That means the point $D=(2,0,0)$ is on the plane. This gives us the displacement vector from the origin of $\vec{d}=\left<2,0,0\right>$.
Let's find a couple more points: If we set $x=z=0$, we the normal form equation becomes $2y=2$, so $P=(0,1,0)$ is on the plane. If we set $x=-1$ and $y=0$, the equation becomes $-1+3z=2$, so $Q=(-1,0,1)$ is on the plane. Therefore $\vec{u}=\vec{DP}=\left<-2,1,0\right>$ and $\vec{v}=\vec{DQ}=\left<-3,0,1\right>$ are vectors in the plane. Hence the vector form of the equation of this plane is $$\vec{r}(s,t)=s\left<-2,1,0\right>+t\left<-3,0,1\right>+\left<2,0,0\right>$$ and the parametric equations are \begin{align} x&=-2s-3t+2\\ y&=s\\ z&=t \end{align}