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

[KingKai]

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.

 

[SammyS]

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.

 

[KingKai]

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.

 

[SammyS]

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.

 

[KingKai]

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?

 

[KingKai]

The Big T represents transpose.

 

[SammyS]

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 .

 

[HallsofIvy]

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:
\left|\begin{array}{ccc}i & j & k \\ -1 & 0 & 2\\a & b & c\end{array}\right|= -2bi+ (2a- c)j- bk
is a normal vector. That, together with the point (1, 1, -1) in the plane gives you the equation.

 

[KingKai]

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:
\left|\begin{array}{ccc}i & j & k \\ -1 & 0 & 2\\a & b & c\end{array}\right|= -2bi+ (2a- c)j- bk
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?

 

[SammyS]

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.

 

[KingKai]

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?

 

[SammyS]

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.

 

[KingKai]

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.