Results 1 to 2 of 2

Math Help - Cartesian equation

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    22

    Cartesian equation

    Hey I am studying Linear and Abstract Algebra and came across this:


    Find the Cartesian equation of the line that passes through the point P(2, 0, -5) and is perpendicular to both the vectors u = (0, 1, 4) and v(-2, -1, 3)



    I am aware that Cartesian equations are in the form: {(x - x0)/a} = {(y - y0)/b} = {(z - z0)/c} , where a,b,c do not equal 0. However, I am not sure how to write the above as a Cartesian equation. If you could help, I thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    Quote Originally Posted by iExcavate View Post
    Hey I am studying Linear and Abstract Algebra and came across this:


    Find the Cartesian equation of the line that passes through the point P(2, 0, -5) and is perpendicular to both the vectors u = (0, 1, 4) and v(-2, -1, 3)



    I am aware that Cartesian equations are in the form: {(x - x0)/a} = {(y - y0)/b} = {(z - z0)/c} , where a,b,c do not equal 0. However, I am not sure how to write the above as a Cartesian equation. If you could help, I thank you!

    The vector form of lines is always easiest.

    To find a vector that's perpendicular to both \mathbf{u} and \mathbf{v} is \mathbf{u}\times\mathbf{v}.


    \mathbf{u}\times\mathbf{v} = \left|\begin{matrix}\phantom{-}\mathbf{i} & \phantom{-}\mathbf{j} & \phantom{-}\mathbf{k} \\ \phantom{-}0 & \phantom{-}1 & \phantom{-}4 \\ -2 & -1 & \phantom{-}3\end{matrix}\right|

     = \mathbf{i}\left|\begin{matrix}\phantom{-}1 & 4 \\ -1 & 3\end{matrix}\right| - \mathbf{j}\left|\begin{matrix}\phantom{-}0 & 4 \\ -2 & 3\end{matrix}\right| + \mathbf{k}\left|\begin{matrix}\phantom{-}0 & \phantom{-}1 \\ -2 & -1\end{matrix}\right|

     = \mathbf{i}[1(3) - 4(-1)] - \mathbf{j}[0(3) - 4(-2) ] + \mathbf{k}[0(-1) - 1(-2)]

     = 7\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}

     = (7, -8, 2).


    Therefore the line that passes through P(2, 0, -5) and perpendicular to both \mathbf{u} and \mathbf{v} is

    (2, 0, 5) + t(7, -8, 2)

     = (2 + 7t, -8t, 5 + 2t).


    So x = 2 + 7t, y = -8t, z = 5 + 2t


    Therefore:

    t = \frac{x - 2}{7} = -\frac{y}{8} = \frac{z - 5}{2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with cartesian equation
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 1st 2010, 02:24 PM
  2. Replies: 4
    Last Post: December 1st 2009, 04:45 PM
  3. Cartesian equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 15th 2008, 07:56 PM
  4. Replies: 4
    Last Post: August 21st 2008, 08:02 AM
  5. Cartesian Equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 26th 2008, 10:09 AM

Search Tags


/mathhelpforum @mathhelpforum