Equation of a plane given point and line in parametric form

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Homework Help Overview

The discussion revolves around finding the equation of a plane given a point P=(-1,0,1) and a line in parametric form represented by r(t)=<3t,t,8>. Participants are exploring the relationship between a point and a vector in the context of plane equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • One participant attempts to use the normal vector and the point to derive the plane equation but expresses confusion about their calculations. Another participant questions the validity of their derived equation and the interpretation of the vector as a line. There is also discussion about the conditions necessary for defining a unique plane.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's approaches. Some guidance has been offered regarding the relationship between points and vectors in defining a plane, but there is no explicit consensus on the correct method or outcome.

Contextual Notes

Participants are navigating the constraints of the problem, including the implications of having only a point and a vector, which may not uniquely determine a plane. There is also mention of potential errors in the derived equations and the need to verify if the given point lies within the proposed plane equations.

musicmar
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Homework Statement


Find the equation of the plane that contains P=(-1,0,1) and r(t)=<3t,t,8>



Homework Equations





The Attempt at a Solution



n * <r-r0>=0
n * <t+2, 2t, 3t> = 0

I distributed the n, adding the terms and obtained:

1/t = -2n(n+2n+3n)

Clearly, I've done something wrong. If someone could point me in the right direction with even how to start this problem correctly, that would be great.

Thanks!
 
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Ok, I set t=1, found a point P0 (2,2,2) and found two vectors OP and OP0 and took the cross product.
I then took the dot product of the normal vector I found above and the given point P.

So, my equation of the plane is 2x + 2x = -1

Can someone tell me if this is valid/and/or correct?
 
musicmar said:

Homework Statement


Find the equation of the plane that contains P=(-1,0,1) and r(t)=<3t,t,8>
You won't be able to get a unique plane if all you know is a point in the plane and a vector that lies in it.
musicmar said:

Homework Equations





The Attempt at a Solution



n * <r-r0>=0
n * <t+2, 2t, 3t> = 0
I'm not following what you're doing above. I get that n is a normal to the plane, but where did <t + 2, 2t, 3t> come from?
musicmar said:
I distributed the n, adding the terms and obtained:

1/t = -2n(n+2n+3n)
?
musicmar said:
Clearly, I've done something wrong. If someone could point me in the right direction with even how to start this problem correctly, that would be great.

Thanks!
 
2x+ 2x= -1?? The plane x= -1/4 is the plane parallel to the yz-plane at x= -1/4.
Did you mean 2x+ 2y= -1 or perhaps 2x+ 2z= -1?

musicmar, a line and a point not on the line determine a plane. But a "vector" is not a "line". It gives the direction but not specific points. Just given a "vector" we might have a line in the direction of the vector through the given point- and that will not determine a plane.

Anyway, to check if 2x + 2x= 4x = -1 is a solution, see if it meets the conditions. Does the given point (-1, 0, 1) lie in it? No, it doesn't; 4(-1)\ne -1. Nor does it lie in 2x+ 2y= -1 or 2x+ 2z= -1. 2(-1)+ 2(0)= -2\ne -1 and 2(-1)+ 2(1)= 0\ne -1.
 

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