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Parametric vector form of cartesian equation

  1. Sep 1, 2015 #1
    How can I find the parametric vector form of a cartesian equation under a specific condition?

    Cartestian equation: $$-2x-y+z=6$$
    I know to find the parametric vector form we can find any 3 points P, Q and R which satisfy the cartesian equation.
    $$ \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} =P \begin{pmatrix} x_2\\ y_2\\ z_2 \end{pmatrix} =Q \begin{pmatrix} x_3\\ y_3\\ z_3 \end{pmatrix} =R $$
    The parametric equation thus becomes:
    $$x=P+\lambda(Q-P)+\mu(R-P)$$
    This part is clear, but my question is how can we find the points P, Q and R if Q-P has been specified?
    $$ \begin{pmatrix} P \end{pmatrix} + \lambda \begin{pmatrix} 1\\ -2\\ 0\\ \end{pmatrix} + \mu \begin{pmatrix} R-P \end{pmatrix} $$
     
  2. jcsd
  3. Sep 1, 2015 #2
    Just rearrange the equation for z, and set x and y to your two paramters.
     
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