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Point Normal Equation Question 
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#1
May2214, 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. 


#2
May2314, 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. 


#3
May2314, 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 slopeintercept 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 2vector ##(\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" onedimensional subset of twodimensional space, is essentially the same as that of a plane, a "flat" twodimensional subset of threedimensional space, can you guess what form the equation of the "flat" (n1)dimensional subset of ndimensional space (called a "hyperplane") has? Edit: Having examined your attachment, I feel the need to mention that the "pointnormal" form of the equation of a plane is comparable to a rearrangement (in a manner similar that that presented above) of the goodold point slope form of the equation of a line. 


#4
May2414, 06:18 PM

P: 68

Point Normal Equation Question



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