SUNY Geneseo Department of Mathematics
Friday, February 17
Math 223
Spring 2023
Prof. Doug Baldwin
(No.)
Based on “Equations for a Plane” in section 1.5.
For example, what is the distance from the plane 2x - y + 3z = 4 to point (1, -2, -1)?
The general idea: think of the general plane equation, and distance to some general point Q with coordinates (x, y, z). The distance from the point to the plane is defined to be the distance along a line perpendicular to the plane.
Since we want a perpendicular distance, it’s distance in the same direction that the plane’s normal points. The normal can be read out of the plane equation.
Finally, imagine any point, P, in the plane, and the vector from that point to Q.
With all of these vectors identified, the distance from Q to the plane is the length of the projection of the PQ vector onto the normal. This length is what the book called the “component of PQ in the direction of n” in the section on dot products, and it gave a formula there too.
Apply it to this problem:
The only thing not explained by the general derivation above is how to find the point P in the plane. Any point will do, as long as its satisfies the plane’s equation. So find the “easiest” triple x, y, z that does so — for example, set x and y to 0 and see what z would have to be, or in this case set y and z to 0 because that leads to a nice simple x = 2, etc. Once you find a point, you calculate the vector PQ and use the component formula:
Notice the negative answer here, which seems strange since distances are always positive. The negative just means that point Q is on the opposite side of the plane from the side the normal points towards. The actual distance is the absolute value of our negative number.
Also notice that it really doesn’t matter what point you pick as P. When we used a different one we still got the same distance:
Which if any of the following planes are parallel? How do you know?
Parallel planes have parallel normals! And parallel vectors are scalar multiples of each other. So find the normals to the planes, and see if there’s ever a constant that you can multiple each component of one of the normals by in order to get another one of the normals.
In this case, none of the planes were parallel (the first and third aren’t parallel either, although the picture doesn’t show that). So could you find a plane parallel to, say, plane 1? Any scalar multiple of plane 1’s normal is a normal to the new plane, so maybe…
But now all the points we can find in the “new” plane are also in the original. And in fact, the new equation is just an alternative equation for the original plane, because I multiplied the constant in the original equation by the same scalar I used for the normal. This illustrates an important idea about planes: their equations aren’t unique, i.e., instead of having “the” equation for a particular plane, you have many.
To find a truly distinct plane that’s parallel to plane 1, use a different constant, for example
Now a point chosen to be in the new plane isn’t in the original, so we know the planes are distinct. This gives you a geometric way to interpret the constant “D” in plane equations: it determines the position of the plane, while the normal determines its orientation.
Problem set 4, on lines and planes, is ready.
Work on it this week, grade it the week after.
But beware that the week after includes Geneseo’s Diversity Summit (February 28), when classes are canceled and we’re all encouraged to participate in the Summit sessions, so I’d prefer not to do meetings that day.
See the handout for more information.
Introduction to vector-valued functions.
Please read “Definition of a Vector-Valued Function” and “Graphing Vector-Valued Functions” in section 2.1 of the textbook.
We’ll look at graphing vector-valued functions with Mathematica, so bring computers to class if you want to try it as we talk about it.