# Normal form of a vector

I had a maths test today and there was a similar question to this: Let $\vec{v}=2\vec{i}+\vec{j}+t(2\vec{i}-\vec{j})$, where t gets the values of all real numbers. Present $\vec{v}$ in the form ax+by+c=0.

I was stuck with the problem of eliminating the letter t. I found out that the direction vector of

$ax+by+c$ is $b\vec{i}+a\vec{j}$.

I wrote $\vec{v}=2\vec{i}+\vec{j}+2t\vec{i}-t\vec{j}=(2+2t)\vec{i}+(1-t)\vec{j}$

Is $b=(2+2t)\vec{i}$

and

$a=(1-t)\vec{j}$

How do I go on from there? Any help would be appreciated.

As you have done, write the vector equation as $$\vec v=x \vec i + y \vec j=\vec i(2+2t) + \vec j (1-t)$$
This means $$\begin {cases} x=2+2t\\ y=1-t \end{cases}$$
find $t$ from one equation and substitute in the other.
There are a couple of ways to find this equation that don’t explicitly involve an algebraic manipulation to eliminate $t$.
One way is to use the point-normal form of equation for a line: $\vec n\cdot(\vec x-\vec p)=0$. Here, $\vec n$ is a vector normal to the line, i.e., perpendicular to its direction vector, and $\vec p$ is any known point on the line. From the vector parametric equation that you’ve been given, the line’s direction vector is $2\vec i-\vec j$, so you can use $\vec n=\pm(\vec i+2\vec j)$. By setting $t$ to zero, you can see that $\vec p=2\vec i+\vec j$ is a point on the line, leading to the equation $$(\vec i+2\vec j)\cdot(x,y)-(\vec i+2\vec j)\cdot(2\vec i+\vec j) = x+2y-4 = 0.$$
Another way uses homogeneous coordinates, in which the vector $\mathscr l = [a:b:c]$ that represents a line corresponds directly to the equation $ax+by+c=0$ via the homogeneous equation $\mathscr l\cdot\mathbf x = 0$. Given two points $\mathbf p$ and $\mathbf q$, the unique line through them is $\mathscr l = \mathbf p\times\mathbf q$. (This is a vector that’s orthogonal to both $\mathbf p$ and $\mathbf q$, hence $\mathscr l\cdot\mathbf p = \mathscr l\cdot\mathbf q = 0$.) The given parametric vector equation of the line provides the two needed points: $[2:1:1]$, which is $\vec p$, above, converted to homogeneous form, and $[2:-1:0]$, which is the point at infinity that corresponds to the line’s direction vector. We compute $[2:1:1]\times[2:-1:0] = [1:2:-4]$, which corresponds to the Cartesian equation $x+2y-4=0$ as derived above.