Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By Plato

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

  1. #1
    Newbie
    Joined
    Mar 2017
    From
    Australia
    Posts
    23

    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.
    Attached Thumbnails Attached Thumbnails Converting parametric equation of a line into Cartesian equation.-q4.png   Converting parametric equation of a line into Cartesian equation.-a4.png  
    Last edited by nalo6451; Apr 18th 2017 at 05:00 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,267
    Thanks
    2641
    Awards
    1

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

    Quote Originally Posted by nalo6451 View Post
    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: $\dfrac{x-2}{-1};~y=3;~z=-1$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2017
    From
    Australia
    Posts
    23

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,267
    Thanks
    2641
    Awards
    1

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

    Quote Originally Posted by nalo6451 View Post
    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.
    Thanks from nalo6451
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Aug 24th 2015, 06:47 PM
  2. Converting a polar equation to Cartesian form.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jul 9th 2011, 05:44 PM
  3. Replies: 4
    Last Post: Dec 1st 2009, 03:45 PM
  4. Replies: 2
    Last Post: Jan 25th 2009, 02:19 PM
  5. Replies: 4
    Last Post: Aug 21st 2008, 07:02 AM

/mathhelpforum @mathhelpforum