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.
Best Answer
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.