Let E be a sphere of radius R with center O. Suppose F is a plane at distance d from O.
If line OC is perpendicular to plane F, the cross-section of the figure by a plane through OC looks like the last figures in the Feb 3 Sheet on Law of Cosines. Using the notation of the figure on the left there, let point Z the the point labeled (a3)I. Then triangle OAZ is a right triangle with sides OA = R, OZ = d, and AZ = r = sqrt(R2 - d2). This distance r is the same for any point A on the intersection of the plane and the sphere. Since Z is also in plane F, we have the intersection is the set of points at distance r from Z, so the set is a circle in plane F with center Z and radius r.
Special cases: If d > R, there is no intersection. If d = R, then r = 0 and the intersection is only only one point, Z = C. Thus the plane is tangent at C if d = R. If d = 0, then Z = O and r = R and the circle is a great circle.
Angle ZOP = arc cos d/R. Is is also arctan r/d and also arcsin r/R.
The right triangles OZA and OAW share an acute angle, so they are similar by AA. Thus we have equal ratios of corresponding sides OW/OA = OA/OZ. Thus OW/R = R/d and OW = R2/d.
Show your work for all parts. Make drawings as needed.
Review the vector concept of dot product and how the equation of a plane Ax+By+Cz = D can be written using dot product. In particular review the fact that the vector [A, B, C] is orthogonal to the plane.
Let E be the sphere of radius 1 with center (0,0,0).
This is the equation of a plane and the plane passes through O = (0,0,0), since 0+0+2*0 = 0.
The line through the poles is the line through O and (1,1,2), so the points are of the form (1-t)O + t(1, 1, 2) = t(1, 1, 2). The points on the sphere are the points with length =1, i.e., the points with x2 +t2+z2 = 1. This says t2(1+1+4) = 1, or t = sqrt(1/6). So the poles are N = sqrt(1/6)(1, 1, 2) and -N = -sqrt(1/6)(1, 1, 2).
Another way to see this uses the dot product. If M = (1, 1, 2), then the equiatio of the plane is M.P = 0, where P = (x,y,z). Then |M| = sqrt M.M = sqrt 6, so N = + or - (1/|M|)M.
The equations both look like x+y+2z = k, where k is chosen so that the distance to 0 is 1, which is the same as saying that the plane passes through N or -N.
A plane passes through N if N is a solution of the equation: This means that sqrt(1/6)(1*1+1*1 +2* 2) =k or -k. Thus k = sqrt 6 or = -sqrt 6.
Answer: x+y+2z = sqrt 6 is one plane and x+y+2z = -sqrt 6 is the other.
Note: This equation is simply N.P = +1 or -1, which makes sense, since N has length 1.
In class and in Assignment 5A we developed the formula
P = (1/(1+ u2+v2))[2u, 2v, u2 + v2 - 1]
for taking a point (u, v, 0) by (inverse) stereographic projection from the plane to the sphere. In class we also solved this problem for a general plane Ax+By+Cz = D. This is an example.
Taking this point P as (x,y,z) and substituting into the equation of the plane, we get
(1/(1+ u2+v2))(2u + 2v + 2(u2 + v2 - 1)) = 1
2u + 2v + 2(u2 + v2 - 1) = 1+ u2+v2
u2+v2 + 2u + 2v -3 = 0
This is the equation. By completing the square we can also write this in a way that shows the center and the radius.
(u+1)2+(v+1)2 = 5
Let E be the sphere of radius R with center (0,0,0).
The directions of the vertices are (1, 1, 1), (1, 1, -1) etc. The points in this direction on the sphere are obtained by dividing by sqrt 3 as was done in 5-3. So the vertices are (1/sqrt 3)(1, 1, 1), (1/sqrt 3)(1, 1, -1), etc. More precisely (and more pedantically) the vertices are the points (1/sqrt 3)(u, v, w), where each u, v, w is +1 or -1 for all 8 possibilities.
One such plane contains the points (1,1,1) and (1,1,-1). By inspection, the plane 1x - 1y +0z = 0 is the equation of the plane. This can also be written as x - y = 0. Likewise, the plane containing (-1, 1, 1) and (-1, 1, -1) has equation x+y = 0.
There are 6 great circles. Each of the 12 edges of the cube has endpoints in a great circle, but the endpoints of the opposite edge also lie on the same circle, so the number of circles = 12/2 = 6. Here are the 6 equations.
x - y = 0, y - z = 0, z - x = 0, x+y = 0, y+z = 0, z + x = 0
The perpendicular bisector of the edge form (1,1,1) to (1, 1, -1) is perpendicular to the edge line, which is parallel to vector (1,1,1) - (1, 1, -1) = (0, 0, 2), so the plane is perpendicular to the z axis, so it has equation z = k. Since the center O is equidistant from (1,1,1) and (1, 1, -1), the point O is on the plane, so k = 0 and the equation is z = 0 (or 0x+0y+1z =0 if you want it to be clearly an equation in 3 variables).
Let E be a sphere of radius R. Suppose that the surface area of E is A (there is a formula but we don't need it for what we will do here).
Example: What is the area of a hemisphere as a fraction of A? (Answer = (1/2)A)
In degrees the area = (t/360)A. In radians, the area = (t/2pi)A
The area is one-half the area above.
In degrees the area = (t/720)A. In radians, the area = (t/4pi)A
The angles are two right angles and the angle t, so the sum is one straight angle + t. In degrees this is 180+t. In radian it is pi + t.
Note after class Wednesday. Notice that this is a special case of the result you derived in class: Girard's Theorem that relates the area of a spherical triangle to its area.