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    Plane Scalar
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Homework Help Overview

The discussion revolves around finding the scalar equation of a plane defined by two points, P(1,1,-1) and Q(0,1,1). Participants explore the implications of having only one direction vector and the resulting infinite number of planes that can pass through the specified points.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the vector equation of the plane and the role of the normal vector. There is an exploration of how to derive the scalar equation given the constraints of the problem. Questions arise regarding the uniqueness of the plane and the implications of having only one direction vector.

Discussion Status

Some participants have provided insights into the relationship between the normal vector and the direction vectors in the plane. There is ongoing exploration of how to express the scalar equation based on the conditions provided, with various interpretations being discussed. No explicit consensus has been reached, but productive lines of reasoning are being explored.

Contextual Notes

Participants note that the scalar equation of the plane is typically expressed in the form ax + by + cz = d, and there is a recognition of the need to ensure that the normal vector is not included in the plane itself. The discussion includes considerations of how to handle the infinite possibilities of the second direction vector.

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Homework Statement



Find the Scalar Equation of a Plane containing the points

P(1,1,-1)
Q(0,1,1)

Homework Equations



ax + by+ cz = d

The Attempt at a Solution



PQ = [-1,0,2]T

[x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[a,b,c]T

^ This is the vector equation.
 
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KingKai said:

Homework Statement



Find the Scalar Equation of a Plane containing the points

P(1,1,-1)
Q(0,1,1)

Homework Equations



ax + by+ cz = d

The Attempt at a Solution



PQ = [-1,0,2]T

[x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[a,b,c]T

^ This is the vector equation.
Hello KingKai. Welcome to PF !

First of all, you should know that this plane is not unique --- there are an infinity of planes passing through these two points.

The vector [a,b,c]T is normal (perpendicular) to the plane, it's not in the plane.
 
Correct me if I am wrong, but the normal to a plane is determined by taking the cross product of the two given direction vectors in the vector equation of the plane. The normal to a plane is not included in the vector equation at all.

Since only 1 direction vector is given in the problem statement, then it is correct that there would be infinitely many planes that satisfy these conditions. That being said, I would like to obtain the scalar equation of this plane and have difficulty doing so.
 
KingKai said:
Correct me if I am wrong, but the normal to a plane is determined by taking the cross product of the two given direction vectors in the vector equation of the plane. The normal to a plane is not included in the vector equation at all.

Since only 1 direction vector is given in the problem statement, then it is correct that there would be infinitely many planes that satisfy these conditions. That being said, I would like to obtain the scalar equation of this plane and have difficulty doing so.
You gave the scalar equation of the plane as ax + by + cz = d .

Then the vector [a,b,c] is normal to the plane. Therefore, it should not be in the vector equation the way you have it.
 
Okay, then if I modify the vector equation to

[x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[x0,y0,z0]T

and the scalar equation of the plane is still in form

ax + by + cz = d

Then how, with the above posted conditions, would I attain a scalar equation for this plane?

The main reason that I am confused is because only 1 direction vector is given to me, the other has infinite possibilities, so how do I derive a scalar equation that satisfies these conditions?
 
The Big T represents transpose.
 
KingKai said:
Okay, then if I modify the vector equation to

[x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[x0,y0,z0]T

and the scalar equation of the plane is still in form

ax + by + cz = d

Then how, with the above posted conditions, would I attain a scalar equation for this plane?

The main reason that I am confused is because only 1 direction vector is given to me, the other has infinite possibilities, so how do I derive a scalar equation that satisfies these conditions?
If you mean that the arbitrary point (x0,y0,z0) is to be in the plane, then the vector equation would be

[x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[x0-1,y[SUB ]0[/SUB]-1,z0+1]T .

However, (besides the fact that this is not a scalar equation) this doesn't ensure that the point (x0,y0,z0) is not on the line determined by points P & Q.

Remember, vector [a,b,c]T must be perpendicular to vector [-1,0,2]T . That implies that the scalar product of these two vectors must be zero. That will give you a relationship among the quantities a, b, and c.

You will get the same relationship is you plug the coordinates of P and then Q into the scalar equation, ax + by+ cz = d .
 
KingKai said:

Homework Statement



Find the Scalar Equation of a Plane containing the points

P(1,1,-1)
Q(0,1,1)

Homework Equations



ax + by+ cz = d

The Attempt at a Solution



PQ = [-1,0,2]T

[x,y,z]T = [1,1,-1] + s[-1,0,2]T + t[a,b,c]Tk
^ This is the vector equation.
If you mean that you want a scalar version of this then you have that [-1, 0, 2] and [a, b, c] are vectors in the plane. Their cross product:
[tex]\left|\begin{array}{ccc}i & j & k \\ -1 & 0 & 2\\a & b & c\end{array}\right|= -2bi+ (2a- c)j- bk[/tex]
is a normal vector. That, together with the point (1, 1, -1) in the plane gives you the equation.

 
HallsofIvy said:
If you mean that you want a scalar version of this then you have that [-1, 0, 2] and [a, b, c] are vectors in the plane. Their cross product:
[tex]\left|\begin{array}{ccc}i & j & k \\ -1 & 0 & 2\\a & b & c\end{array}\right|= -2bi+ (2a- c)j- bk[/tex]
is a normal vector. That, together with the point (1, 1, -1) in the plane gives you the equation.


So, following this, I take the normal vector

n = [-2b, 2a-c, -b]T

and the point P(1,1,-1)

and compile the scalar equation

-2b(x-1) + (2a-c)(y-1) -b(z+1) = d

-2bx + 2b +2ay -2a -cy +c -bz -b = d


Would this be correct?
 
  • #10
KingKai said:
So, following this, I take the normal vector

n = [-2b, 2a-c, -b]T

and the point P(1,1,-1)

and compile the scalar equation

-2b(x-1) + (2a-c)(y-1) -b(z+1) = d

-2bx + 2b +2ay -2a -cy +c -bz -b = d

Would this be correct?
If your scalar equation for the plane is ax + by + cz = d, then the vector [a, b, c]T is not in the plane. That vector is normal to the plane.

Like I suggested, plug (1,1,-1) into ax + by + cz = d. Then plug (0,1,1) into ax + by + cz = d.

That gives a + b - c = d and b + c = d .

That gives a relation between a & c, so you can eliminate one of them.
 
  • #11
SammyS said:
If your scalar equation for the plane is ax + by + cz = d, then the vector [a, b, c]T is not in the plane. That vector is normal to the plane.

Like I suggested, plug (1,1,-1) into ax + by + cz = d. Then plug (0,1,1) into ax + by + cz = d.

That gives a + b - c = d and b + c = d .

That gives a relation between a & c, so you can eliminate one of them.



a + b -c = d b + c = d
c = d - b

Sub c = d - b into

a + b - c = d


a + b - (d - b) = d

a + b -d + b = d

a + 2b = 2d From This the scalar equation would be:

ax +2by = 2d True?
 
  • #12
KingKai said:
a + b -c = d b + c = d
c = d - b

Sub c = d - b into

a + b - c = d


a + b - (d - b) = d

a + b -d + b = d

a + 2b = 2d From This the scalar equation would be:

ax +2by = 2d True?

Where do z go ?


If a + b - c = d and b + c = d, then a + b - c = b + c → a = 2c.

Using that a = 2c and b + c = d gives:

2cx + by + cz = b+c .

You can play around with this to make it look nicer.
 
  • #13
Thank You SammyS for all your help!

I'm having trouble typing out matrices on this forum, is there a thread you can refer me that will teach me the syntax of how I can type out proper matrices the way HoI did above?

I have another question from my linear algebra textbook that I need help with but I can't type the matrix.
 

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