Point Normal Equation Question

In summary, the equation ##ax +by +cz +d = 0## represents a plane orthogonal to the vector ##(a,b,c)## in three-dimensional space. The constant term ##d## determines the position of the plane away from the origin. This can be compared to the equation of a line in two-dimensional space, where the constant term determines the position of the line away from the origin. The "point-normal" form of the equation is comparable to rearranging the point-slope form of a line.
  • #1
Regtic
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##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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  • #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.
 
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  • #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.
 
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  • #4
gopher_p said:
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.
 
  • #5


The d in the equation represents the distance of the plane from the origin. This is known as the offset or intercept term. It is necessary to fully define the position of the plane in three-dimensional space. Without the d term, the equation would only represent a family of parallel planes, all perpendicular to the vector (a,b,c). The d term allows for a specific plane to be defined within that family.

Geometrically, the d term determines where the plane intersects the z-axis. If d is positive, the plane will intersect the z-axis above the origin, and if d is negative, it will intersect below the origin. This can also be seen when graphing the equation in three-dimensional space, as the d term shifts the position of the plane along the z-axis.

In summary, the d term in the point normal equation is crucial for fully defining the position of the plane in three-dimensional space and determining its relationship to the origin.
 

1. What is the Point Normal Equation?

The Point Normal Equation is a mathematical formula used to find the perpendicular distance from a point to a plane. It is also known as the shortest distance formula.

2. How is the Point Normal Equation derived?

The Point Normal Equation is derived from the dot product of two vectors - the vector from the point to a point on the plane, and the vector perpendicular to the plane. This dot product is set equal to zero, and the resulting equation is solved for the distance.

3. When is the Point Normal Equation used?

The Point Normal Equation is used in various fields, including geometry, physics, and engineering. It is commonly used to find the shortest distance between a point and a plane, which can be useful in calculating distances or determining intersections between objects.

4. Can the Point Normal Equation be used in three-dimensional space?

Yes, the Point Normal Equation can be used in three-dimensional space. In this case, the equation will have three variables representing the coordinates of the point and three variables representing the coefficients of the plane.

5. Are there any limitations to the Point Normal Equation?

Yes, the Point Normal Equation can only be used to find the distance between a point and a plane. It cannot be used to find the distance between two points or between two planes. Additionally, it assumes that the point is not on the plane, otherwise the distance would be zero.

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