Equation of Plane: Solution + Finding Vector

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Homework Help Overview

The discussion revolves around finding the equation of a plane and the associated normal vector. Participants express concerns about the steps taken to derive the equation without explicitly identifying the normal vector.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Some participants question how the equation of the plane was derived without detailing the normal vector. Others suggest that the normal vector can be used to find the equation, while also discussing the form of the equation and the role of intercepts in determining coefficients.

Discussion Status

Participants are actively exploring different interpretations of the problem, with some providing insights into the relationship between the normal vector and the equation of the plane. There is a recognition of the need for clarity regarding the steps involved in the derivation.

Contextual Notes

There is mention of a paint document containing the original question and solution, which may lack certain details about the derivation process. Participants are also discussing the arbitrary nature of one of the constants in the plane equation.

Miike012
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The question and solution is in the paint doc. My concern was how did they find the equation of the plane without finding the vector normal to the plane? I am guessing they found the vector but left out the steps...
 

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They could find the normal vector, and got the equation of the plane using it.

ehild
 
Last edited:
Miike012 said:
The question and solution is in the paint doc. My concern was how did they find the equation of the plane without finding the vector normal to the plane? I am guessing they found the vector but left out the steps...
The equation for a plane can be given in the form, [itex]ax+by+cz=d\ .[/itex] Of course, one of those constants is arbitrary.

Just plug-in the each set of coordinates for the three intercepts, individually, to find [itex]\displaystyle \frac{a}{d}\,,\ \frac{b}{d}\,\ \text{ and },\ \frac{c}{d}\,,\[/itex] then find a convenient value to use for d.
 
SammyS said:
The equation for a plane can be given in the form, [itex]ax+by+cz=d\ .[/itex] Of course, one of those constants is arbitrary.

Just plug-in the each set of coordinates for the three intercepts, individually, to find [itex]\displaystyle \frac{a}{d}\,,\ \frac{b}{d}\,\ \text{ and },\ \frac{c}{d}\,,\[/itex] then find a convenient value to use for d.

That is much easier... :cool:

ehild
 
wouldn't it be a = d/x, b = d/y, and c = d/z where (x,y,z) are points on the plane and <a,b,c> is the normal vector
hence if (x,y,z) = (0,0,z) then cz = d and c = d/z ... you can do the same for a and b components of the normal.
 
Miike012 said:
wouldn't it be a = d/x, b = d/y, and c = d/z where (x,y,z) are points on the plane and <a,b,c> is the normal vector
hence if (x,y,z) = (0,0,z) then cz = d and c = d/z ... you can do the same for a and b components of the normal.

For the x-intercept, x=1, y=0, and z=0.

Therefore, a(1) + 0 + 0 = d  →  a/d =1,  etc.
 

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