## plane whose Normal is (0, 0, 0) not possible, right? (3)

### 1 Name: Anonymous Scientist : 2006-11-09 06:22 ID:8bYyLTo1

In mathematics, is a plane whose Normal is (0, 0, 0) not possible, or "undefined"? I would think so since its normal vector isn't pointing to anywhere at all for it to be perpendicular to.

I just saw an equation for collision where the result would be a divide-by-zero because you'd take the dot-product of a plane's normal and a velocity vector for the divisor in the equation and it occured to me that if either the velocity is motionless (which you check for before doing this equation) or if the plane's normal is all zeros then the dot product would equal 0 and you'd get a divide-by-zero so... I'm guessing... a plane with a (0, 0, 0) normal (regardless of its D value) just doesn't exist...

### 2 Name: Anonymous Scientist : 2006-11-09 09:44 ID:9g5vx6Oq

I'm not going to dig out my vector calc book, but I don't see any problem with this. <0,0,0> has no length -- how could anything, therefore, be normal to it?

### 3 Name: Anonymous Scientist : 2006-11-09 10:49 ID:MfUH+pDa

The definition of a plane using the normal (a,b,c) is a*x + b*y + c*z + d = 0. You easily see that if (a,b,c) is (0,0,0), then this will be true for all points if d is 0, and true for no points if d is not 0.

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