Vector Planes & Orthogonality - Help

[mrneglect]

Vector Planes & Orthogonality -- Help!

I must be doing something really stupid, and I'll kick myself when you point it out, but I'm having difficulty with this question:

Find the unit normal to the plane x + 2y – 2z = 15. What is the distance of the plane from the origin?

OK, so I know I need three points in the plane to identify it. I found points:

A = (15,0,0)
B = (15,1,1)
C = (13,2,1)

They all seem to satisfy the equation, right?

So now I find two nonparallel vectors in this plane:

A->B = 0i + 1j + 1k
A->C = -2i + 2j + 1k

They do indeed join those points, right?

So now I want to find an orthogonal vector, so I take the cross product by calculating the determinant of the 3x3 matrix:

[i j k]
[0 1 1]
[-2 2 1]

= (1-2)i - (-2-0)j + (0+2)k
= -i + 2j + 2k

This vector ought to be orthogonal to vectors A->B, A->C and B->C (and their negatives). But I try to test it with the usual method of adding up the parts and seeing if they result in zero, using A->C as my comparison vector:

(-2 * -1) + (2 * 2) + (1 * 2) = 8

Why doesn't this equal zero? I have a feeling that the middle (j) term is wrong, because if that was (2 * -2) then the whole thing would equal zero, but I don't see where I've got my signs wrong in the method.

Any help would be much appreciated.

Cheers!

 

[slider142]

[i j k]
[0 1 1]
[-2 2 1]

= (1-2)i - (-2-0)j + (0+2)k

The second term in the sum should be -(0 - (-2))j = -2j.

 

[mrneglect]

Oh, I see, so it doesn't wrap around! That's probably why it's "+ i - j + k". Thanks for that. I knew it was something simple.

 

[HallsofIvy]

Actually, you should have learned early that if a plane is given by Ax+ By+ Cz= D, the <A, B, C> is normal to the plane. You don't need to do all that work. Since the plane is given as x + 2y – 2z = 15, you know that a normal vector is <1, 2, -2>.

 

[mrneglect]

Well for some reason that's not in our course, but I shall certainly be using it from now on!