Find Cartesian equation of plane containging line with parametric equations and point

  • Thread starter craka
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  • #1
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Homework Statement


Question is
"The Cartesian equation of the plane containing the line x=3t , y =1+t , z=2-t and passing through the point (-1,2,1) is?"


Homework Equations



[tex]
\begin{array}{l}
n \bullet (r - r_0 ) = 0 \\
< n_1 ,n_2 ,n_3 > \bullet < x - x_0 ,y - y_0 ,z - z_0 > = 0 \\
\end{array}
[/tex]


The Attempt at a Solution



direction vector is < 3 , 1, -1>

[tex]
\begin{array}{l}
< - 1,2,1 > \bullet < x - 3,y - 1,z + 1 > = 0 \\
- x + 2y + z = 2 \\
\end{array}
[/tex]

But this doesn't appear to be right. Could someone help me out here please. I'm at a lost on how to do this. Thanks
 

Answers and Replies

  • #2
arildno
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1. Pick a point ON the line, (f0,1,2) for example, and set up the equation governing this.
You have three unknowns (essentially, the components of the normal vector).
Utilize the fact that irrespective of the value of t, the whole line should be included in the plane.
This will give you a single equation for the three unknowns.
2. Also require that the given solitary point should lie in the plane.
This will give you the second equation for your three unknowns.

This is what you need, you'll end up with a free scaling parameter for the normal (i.e, its length), as you should.
 
  • #3
HallsofIvy
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Or: Choose two points on the line, p0 and p1. Determine the vector from p0 to p1 and the vector from p0 to (-1, 2, 1). Take the cross product of those two vectors to find the normal to the plane.
 
  • #4
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I still haven't been able to do this.

I tried to do with cross product of <0,1,2> x <-1,2,1>
to get vector normal , which was <-3,-2,1>

than did <-3,-2,1> . < x- (-1), y-2, z-1 >=0

which worked out to be -3x-2y+z=0 however this is not the answer still.
 

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