Which Plane Passes Through the Intersection Line and Satisfies Given Conditions?

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Discussion Overview

The discussion revolves around finding the scalar equation of a plane that passes through the intersection of two given planes and a specific point, while also satisfying certain distance conditions from the origin and another point. The scope includes mathematical reasoning and problem-solving related to geometry and linear algebra.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant seeks to find the equation of a plane that meets specific conditions, including being 2 units from the origin and 3 units from the point (5, -3, 7).
  • Another participant requests clarification on what the original poster has attempted and where they are encountering difficulties.
  • A participant explains their approach, including finding a normal vector through the cross product of the normals of the given planes and identifying points on the intersection line.
  • Some participants express confusion regarding the relevance of the distance conditions to the problem, questioning how they relate to the unique plane that passes through the specified point.

Areas of Agreement / Disagreement

There is no consensus on the relevance of the distance conditions to the problem, with some participants questioning their necessity while others focus on the mathematical approach to finding the plane.

Contextual Notes

Participants have not fully resolved the implications of the distance conditions, and there may be assumptions regarding the uniqueness of the plane that have not been explicitly stated.

SSUP21
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Find the scalar equation of the plane which passes through the line intersection of planes x+y+z-4=0 and y+z-2= 0 that goes through (2,4,7) and satisfies the conditions
a) it is 2 units from the orgin
b) it is 3 units from the point a(5,-3,7)

I would really apprecaite if some toke me through the steps on how the solve the problem

Thank You
 
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Hi SSUP21! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
Hi
and thanks for replying for this one i knew that to find a plane i needed a normal and a point
the intersection of the given planes is a vector in direction of our plane so i cross product their normals

i need another vector i had a point i needed another point so I assume that any point on the intersection of the two give planes is also on the third plane so let y = 0 and solved to find the point
then i found my second vector by doing cross product of the two point
then substituted in our given point to solve for D in equation ax + by+ cz + D=0

i have my plane but I am stuck at trying to satisfy the two conditions
 
Hi SSUP21! :smile:
SSUP21 said:
Hi
and thanks for replying for this one i knew that to find a plane i needed a normal and a point
the intersection of the given planes is a vector in direction of our plane so i cross product their normals

Yes, that's absolutely right. :smile:
i need another vector i had a point i needed another point so I assume that any point on the intersection of the two give planes is also on the third plane so let y = 0 and solved to find the point
then i found my second vector by doing cross product of the two point
then substituted in our given point to solve for D in equation ax + by+ cz + D=0

i have my plane but I am stuck at trying to satisfy the two conditions

mmm … I don't really follow what you're doing. :redface:

In fact, looking at the question more carefully, I don't really understand that either …
Find the scalar equation of the plane which passes through the line intersection of planes x+y+z-4=0 and y+z-2= 0 that goes through (2,4,7) and satisfies the conditions
a) it is 2 units from the orgin
b) it is 3 units from the point a(5,-3,7)

there's only one plane through that line that goes through (2,4,7), so what do a) and b) have to do with it?? :confused:
 

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