Question

IN point normal form, how do you know if it is an equation of aline or a plane

Answer 1

I think it depends on how the functions is stated. Such as f(x) being a line and f(x,y) being a plane. I hope that helps.

Answer 2

Point normal form defines a set of points on a plane by giving the normal vector to the plane and two points lying on the plane. For example, if we have the normal vector N and points on the plane a and b, point normal form gives the equation of the plane as:

N ⋅ (a - b) = 0 (⋅ is the dot product operation).

Remember that the dot product of N and the vector ba is 0 because ba lies in the plane and thus, like all vectors in the plane, is perpendicular to N. If N ⋅ (a - b) ≠ 0, then the equation does not define a plane. In the case that the equation defines a line, vector N and vector ba will be parallel. If N ∥ ba, then N ⋅ (a - b) = 1 (from the definition of the dot product: A ⋅ B = ∣A∣∣B∣cosθ.)

Note that the equation N ⋅ (a - b) = 1 defines a line if and only if points a and b are on the vector N (or the vector cN where c is a scalar constant).

The answer comes down to this, an equation in point normal form offers four general possibilities when N is any given vector and a and b are any two given points.

1) N ⋅ (a - b) = 0 ; defines a plane. N is the normal vector to the plane.

2) N · (a - b) ≠ 0 ; may define a plane. If so, N is not the normal vector to this plane.

3) N ⋅ (a - b) = 1 ; defines a line.

4) N ⋅ (a - b) = 1 ; does not define a line.

Case #1: As long as this equation is true, it will always define a plane. N will be the normal vector and points a and b will lie on the plane.

Case #2: Here, N and ab are not perpendicular, thus N is not the normal vector to the plane. Any two vectors define a plane, thus this equation defines a plane unless...

Case #3: Vector N and vector ab are parallel. In some cases, two parallel vectors can define a plane, but the equation of such a plane would look very different. Because we said this equation defines a line, a and b must be points on vector N (or on a vector cN, in other words, any vector that goes in the same direction as N.)

Case #4: In this case N and ab are parallel, but a and b are not on vector cN. Thus this equation does not define a line (or any set of continuous points in a conventional algebra).

I hope that helps. Please correct me if you think I've made a mistake. Best of luck!

Answer 3

Point-normal form can also represent a line if it's vectors are 2D instead of 3D.
Consider:
<nx, ny>*<x-x0, y-y0> = 0
=> nx(x-x0) + ny(y-y0) = 0
=> ny(y-y0) = -nx(x-x0)
=> y-y0 = -(nx/ny)*(x-x0)

You might recognize the above equation as the point-slope form of a line ( http://www.purplemath.com/modules/strtlneq2.htm). The equation here represents a line with slope -nx/ny that travels through the point <x0, y0>