hey all,
Having problems trying to find a parametric equation of the plane that is parallel to the plane 3x + 2y -z =1 and passes through the point P(1,1,1).
I have tried almost all possible (wrong)ways of solving this question. Thanks
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hey all,
Having problems trying to find a parametric equation of the plane that is parallel to the plane 3x + 2y -z =1 and passes through the point P(1,1,1).
I have tried almost all possible (wrong)ways of solving this question. Thanks
The plane $\displaystyle 3x+2y-z=4$ is parallel to the given plane and contains the point $\displaystyle (1,1,1)$.Quote:
Originally Posted by Oiler
What is the parametric equation of that plane?
The parametric equation for the plane is x=t_1, y=t_2, z=3t_1 + 2t_2-1. Not sure how I can get the vectors that go through the point (1,1,1)
Given any plane $\displaystyle Ax+By+Cz=D$ and point $\displaystyle P: (p,q,r)$Quote:
Originally Posted by Oiler
then the plane $\displaystyle Ax+By+Cz=Ap+Bq+Cr$ is parallel to the given plane and contains $\displaystyle P$.
Is the problem to "find the plane" or specifically to "find parametric equations for the plane"?
Any plane parallel to Ax+ By+ Cz= D is of the form Ax+ By+ Cz= E and E can be determined by a single point in the plane.
You ca get parametric equations for Ax+ By+ Cz= D by, for example, solving for z: $\displaystyle z= \frac{D}{C}-\frac{A}{C}x- \frac{B}{C}z$ and then using x and y as parameters. Or if you prefer using other letters, say u and v,
$\displaystyle x= u$
$\displaystyle y= v$
$\displaystyle z= \frac{D}{C}-\frac{A}{C}u- \frac{B}{C}v$
