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?
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)
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 =".)
Well, that first one is 0a+ 0b+ 0c= 0 = d so your other two equations areI 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)
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.
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?
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.