# Thread: Converting parametric equation of a line into Cartesian equation.

1. ## Converting parametric equation of a line into Cartesian equation.

Hi all. Just looking for some help on a question. I shall attach question and answer below.

I somewhat know how to convert parametric into Cartesian equations (by just transferring over the constants, and plugging into the formula: x-x0/a = y-y0/b = z-z0/c. When I do the working I get the y=3 and z=-3 correct, but I also get x-2/1, which isn't in the answers.

Someone please tell me where I'm going wrong?

I have a feeling it has something to do with the vector being -i but I'm not sure.

2. ## Re: Converting parametric equation of a line into Cartesian equation.

Originally Posted by nalo6451
Hi all. Just looking for some help on a question. I shall attach question and answer below.

I somewhat know how to convert parametric into Cartesian equations (by just transferring over the constants, and plugging into the formula: x-x0/a = y-y0/b = z-z0/c. When I do the working I get the y=3 and z=-3 correct, but I also get x-2/1, which isn't in the answers.
Someone please tell me where I'm going wrong?
I have a feeling it has something to do with the vector being -i but I'm not sure.
I don't not know what your author expects for a symmetric form. I know what I use,
The line thru the point $(2,3,-1)$ with the direction vector $-\bf{i}$ is

parametric: $\left\{ \begin{array}{l}x=2-t\bf{i}\\y=3\\z=-1\end{array} \right.$

symmetric: $\dfrac{x-2}{-1};~y=3;~z=-1$

3. ## Re: Converting parametric equation of a line into Cartesian equation.

I think I've sort of figured it out. If b=c=0 but a doesn't = 0 then there is no restriction on a, thus a can be any real number. I guess here it's just not explicitly written in the solutions.

Now I just need to figure out how the vector equation is 2i+3j-3k+ti. Not sure where the positive ti comes from considering the vector provided is negative (v = -i).

Again, any help anyone can provide would be appreciated

4. ## Re: Converting parametric equation of a line into Cartesian equation.

Originally Posted by nalo6451
I think I've sort of figured it out. If b=c=0 but a doesn't = 0 then there is no restriction on a, thus a can be any real number. I guess here it's just not explicitly written in the solutions.
I have said that different authors use different terms for concepts.
I here give you an example of what I have written. If you cannot find it use please disregard it totally.
Consider the line through the point $\mathcal{P}: (2,-1,-3)$ having direction $\vec{\bf{v}}=-5\bf{i}+2\bf{k}$.

Vector equation: $\ell(t)=\mathcal{P}+t\vec{\bf{v}}=<2,-1,-3>+t\vec{\bf{v}}=\left<2-5t,-1,-3+2t\right>$

Parametric equation:$\ell(t)=\left\{ \begin{array}{l}x=2-5t\\y=-1\\z=-3+2t\end{array} \right.$

Symmetric form: $\dfrac{x-2}{-5}=\dfrac{z+3}{2}; y=-1$.

I hope that example helps you. If not just forget what I have said. I do not want to confuse you.

It is 11:05 pm in this part of the American South-bedtime, so goodnight.