Homework Help: Scalar Equation of a Plane

1. May 30, 2012

KingKai

1. The problem statement, all variables and given/known data

Find the Scalar Equation of a Plane containing the points

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

2. Relevant equations

ax + by+ cz = d

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 30, 2012

SammyS

Staff Emeritus
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. May 30, 2012

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.

4. May 31, 2012

SammyS

Staff Emeritus
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. May 31, 2012

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?

6. May 31, 2012

KingKai

The Big T represents transpose.

7. May 31, 2012

SammyS

Staff Emeritus
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. May 31, 2012

HallsofIvy

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.

9. Jun 1, 2012

KingKai

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. Jun 1, 2012

SammyS

Staff Emeritus
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. Jun 1, 2012

KingKai

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. Jun 1, 2012

SammyS

Staff Emeritus
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. Jun 1, 2012

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.