How to Type Matrices in a Forum to Proper Syntax and Formatting

In summary: 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 and end up with an equation of the form ax + by + cz = d .In summary, the scalar equation of the plane containing points P(1,1,-1) and
  • #1
KingKai
34
0

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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
The Big T represents transpose.
 
  • #7
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 .
 
  • #8
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.

 
  • #9
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.
 

FAQ: How to Type Matrices in a Forum to Proper Syntax and Formatting

What is a scalar equation of a plane?

A scalar equation of a plane is an equation that describes a plane in terms of its distance from the origin and the direction of its normal vector.

How do you write a scalar equation of a plane?

To write a scalar equation of a plane, you need the coordinates of three points that lie on the plane. You can then use these points to find the normal vector and the distance from the origin, which are used in the equation.

What is the significance of the scalar equation of a plane?

The scalar equation of a plane is useful in many areas of mathematics and science, including geometry, physics, and engineering. It allows us to describe the position and orientation of a plane in a concise and precise way.

Can a scalar equation of a plane be used in three-dimensional space?

Yes, a scalar equation of a plane can be used in three-dimensional space. The equation will have three variables, and the normal vector will have three components.

How is a scalar equation of a plane related to linear algebra?

The scalar equation of a plane is closely related to linear algebra, as it involves finding the normal vector and using it to solve for the distance from the origin. This process is similar to finding the solution to a system of linear equations, which is a fundamental concept in linear algebra.

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