Thread: ngEquation of the plane

I need some help please!

p = 2i=j=2k and q=2i+2j

I am trying to find the equation of the plane OPQ, where O is the origin in R3.

Here is where I am

Let the equation of the plane be

ax + by + cz = d (S.1)

Substituing the coordinates of the three points in turn into the equation (S.1), we obtain

= d (S.2)
2a + b + 2c = d (S.3)
2a + 2b = d (S.4)

Could you please clean up your notation a bit? You have p = 2i = j = 2k; all of those inequalities are false! I'm guessing you mean +'s and -'s in there somewhere. Which ones are which?

Originally Posted by Arron

I need some help please!

p = 2i=j=2k and q=2i+2j

I assume you mean p= 2i- j- 2k and q= 2i+ 2j. Whoever put the "-" key next to the "=" key was an evil person!
(But then you use the coordinates (2, 1, 2) below. Did you mean p= 2i+ j+ 2k? (Of course, "+" is "shift =".)

I am trying to find the equation of the plane OPQ, where O is the origin in R3.

Here is where I am

Let the equation of the plane be

ax + by + cz = d (S.1)

Substituing the coordinates of the three points in turn into the equation (S.1), we obtain

= d (S.2)
2a + b + 2c = d (S.3)
2a + 2b = d (S.4)

Well, that first one is 0a+ 0b+ 0c= 0 = d so your other two equations are
2a+ b+ 2c= 0 and 2a+ 2b= 0. Subtract those two equations and you will eleminate a so you can solve the equation, say, for b in terms of c. Then put that back into the second equation and solve for a in terms of c. Because you can multiply ax+ by+ cz= d by any number and have a new equation for the same plane, you cannot solve for specific numbers. Perhaps assuming c= 1 will let you solve for a and b.

You also, by the way, could do this: the equation of a plane with normal vector Ai+ Bj+ Ck and containing the origin is
Ax+ By+ Cz= 0. And you can get the normal vector by taking the cross product of two vectors in the plane- p and q. Try that as a check.

Sorry I meant p=2i+j+2k.

Originally Posted by Arron

Sorry I meant p=2i+j+2k.

Have you been able to carry through HallsofIvy's solution method with that correction?

Hi, sorry for late reply. After carrying out hallsofivy's solution method I get,

-x+2y+z=0

Is this correct?

I'm afraid not. Your solution contains the origin, but not either of the other two points. Double-check your algebra, or post it online.

After substituting 0 for d I get

2a + b +2c =0
2a + 2b = 0

Take away one from the other I get

b + 2c = 0

that is

b = -2c

put this back into the second equation I get

2a -2c = 0

That is

a = c

The equation of the plane is

cx - 2cy + cz = 0

That is

x - 2y + z = 0

How about now?

subtracting "one from the other" does not give b + 2c = 0. check your signs.

Deveno beat me to the algebra, so I'll just mention this:

You can check your own work quite easily: your candidate for the plane was x - 2y + z = 0. Well, the plane needs to contain the three points (0,0,0), (2,1,2), and (2,2,0). Plug those coordinates into the plane equation and see if you get zero. You can see that the point (2,1,2) gives you problems.

I got the algebra a bit mixed up

I make the equation of the line

-2x + 2y + z = 0

Thanks for all your help!

You mean the equation of the plane, right? Yep, I'd say you've got it now. You're welcome!