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Point Normal Equation Question

by Regtic
Tags: equation, normal, point
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Regtic
#1
May22-14, 11:02 PM
P: 68
##ax +by +cz +d = 0## is the equation of a plane orthogonal to ##(a,b,c)## but why is there a d in the equation? What does it do to the plane geometrically?

http://i.imgur.com/sMiLhHc.jpg?1

I don't see how the d fits into that geometric interpretation.
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bhillyard
#2
May23-14, 12:48 AM
P: 35
There are an infinite number of parallel planes perpendicular to (a,b,c). Effectively the d determines which one you are considering, sort of how far along (a,b,c) it crosses (a,b,c).
Think of the 2 dimensional case - ax+by+c=0, a & b determine the slope of the line and c determines where it crosses the axes. There would be an infinite number of parallel lines in the plane all described by the a,b combination, the c fixes on a particular one of the lines.
gopher_p
#3
May23-14, 12:56 AM
P: 477
If it's nonzero, ##d## moves the plane away from the origin.

Remember ##y=mx+b## from algebra? For the sake of demonstration, let's change the letters around to ##y=-\frac{\alpha}{\beta}x-\frac{\gamma}{\beta}##, so that the ##m=-\frac{\alpha}{\beta}## and ##b=-\frac{\gamma}{\beta}##. The equation of our line can now be rewritten as ##\alpha x+\beta y+\gamma=0##. The ##\gamma## of the second from is playing a similar role to the ##b## of the slope-intercept form; it's giving us information regarding a shift away from the origin.

If you recall, the slope of the line perpendicular to this one is given by ##m_\perp=-\frac{1}{m}=\frac{\beta}{\alpha}##. And given this information it's not too hard to check that the 2-vector ##(\alpha,\beta)## is normal to the line that we started out with.

Switching back to "normal" letters, we have what's called the "standard form" of the equation of a line given by $$ax+by+c=0$$ immediately comparable to the standard form of the equation of a plane $$ax+by+cz+d=0$$ with all of the letters playing comparable roles; the coefficients of the variable terms tell us what a normal vector is, and the constant term provides a shift away from the origin.

Now having been shown how the equation of a line, "flat" one-dimensional subset of two-dimensional space, is essentially the same as that of a plane, a "flat" two-dimensional subset of three-dimensional space, can you guess what form the equation of the "flat" (n-1)-dimensional subset of n-dimensional space (called a "hyperplane") has?

Edit: Having examined your attachment, I feel the need to mention that the "point-normal" form of the equation of a plane is comparable to a rearrangement (in a manner similar that that presented above) of the good-old point slope form of the equation of a line.

Regtic
#4
May24-14, 06:18 PM
P: 68
Point Normal Equation Question

Quote Quote by gopher_p View Post
If it's nonzero, ##d## moves the plane away from the origin.
This was the closest answer to the question I was really asking. Bhillyard you made a very interesting point that didn't occur to me though, about how the constant is the identifying coordinate of a specific line among an infinite amount of parallel lines. I'm downloading a 3d graphing program to get a better look. Thanks to both of you for your time and help. I appreciate it.


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