Parametric vector form of cartesian equation

  • #1

Main Question or Discussion Point

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?
$$ \begin{pmatrix} P \end{pmatrix} + \lambda \begin{pmatrix} 1\\ -2\\ 0\\ \end{pmatrix} + \mu \begin{pmatrix} R-P \end{pmatrix} $$
 

Answers and Replies

  • #2
667
128
Just rearrange the equation for z, and set x and y to your two paramters.
 

Related Threads on Parametric vector form of cartesian equation

  • Last Post
Replies
9
Views
21K
  • Last Post
Replies
8
Views
753
Replies
10
Views
831
Replies
2
Views
6K
Replies
12
Views
2K
  • Last Post
Replies
1
Views
11K
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
3
Views
716
Top