# Point Normal Equation Question

by Regtic
Tags: equation, normal, point
 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.