1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Point Normal Equation Question

  1. May 22, 2014 #1
    ##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?


    I don't see how the d fits into that geometric interpretation.
  2. jcsd
  3. May 23, 2014 #2
    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.
  4. May 23, 2014 #3
    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.
    Last edited: May 23, 2014
  5. May 24, 2014 #4
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook