We employ the synthetic vector analysis to obtain the centers of the circum and in circles of a triangle. These constructions for the spherical triangle follow the same algorithm as that usually presented for a plane triangle in a synthetic geometry class. Hence, the resulting circum and in centers of the spherical triangle also are the corresponding centers for the plane triangle through the vertices of the spherical triangle. The magnitudes of the corresponding radii, however, are different for the spherical and plane triangles; because we measure distances differently. Surprisingly, finding the magnitude of the radii from the known centers is easy in the spherical case, but difficult in the plane case. In the plane case, we had obtained these magnitudes in a different way; we do not attempt to obtain them here.
We continue our unusual habit of writing the inner (dot) product of a pair of vectors A and B as the juxtaposition AB. Since our type-face does not have a cross, we employ the lower-case 'x' to indicate the cross product A x B.
Definition: The vectors A and B define the plane pA + qB, where (p, q) is in Real x Real.
Lemma: Colinearity. The vector B = pA is collinear to the vector A, where p is in Real.
Proof: For any vector C such that AC is zero, BC also is zero. QED.
Proposition: Normalization. The vector B = A / sqrt(AA) is a unit vector, collinear to the vector A.
Proof: Colinearity follows from the lemma. BB = 1. QED.
Theorem: Orthogonalization. Given the vectors A and B. The vector
C = B - A (AB) / (AA)
is orthogonal to A and in the (A, B) plane.
Proof: AC = 0. Take p = - (AB) / (AA) and q = 1. QED.
Corollary: Orthonormalization. The vector D = C / sqrt(CC) is collinear to C, orthogonal to A, normal, and in the (A, B) plane.
Proofs are obvious.
Theorem: The orthogonal to a pair of vectors. Given three non-coplanar vectors A, B, and C. Then the vector
X = (((BC) (AB) - (AC) (BB)) A + ((AC) (AB) - (BC) (AA)) B) / ((AA) (BB) - (AB) (AB)) + C
is orthogonal to each of A and B. Likewise, the pseudo-vector Y = A x B is orthogonal to each of A and B.
Proof: AX = 0 and BX = 0. Remember that the inner (dot) product is commutative. AY = 0 and BY = 0; because the box-product with any vector repeated is zero, as shown presently. QED.
Definitions: A plane (B, C) is said to be polar to a vector A iff (= if and only if) the vectors A and B are orthogonal and the vectors A and C also are orthogonal. The vector A is the pole of the plane (B, C).
Lemma: In a positive-definite metric space, a vector is zero (null) iff (= if and only if) its magnitude is zero.
Proof is obvious.
Proposition: The magnitude of the cross-product A x B is the area of the parallelogram defined by the vectors A and B. Hence the cross-product of two vectors is zero iff (= if and only if) they are collinear.
Proofs are obvious.
Definition: The box product, written ABC of the vectors A, B, and C is
ABC = A(B x C)
Corollary: The box product ABC is the volume of the parallelopepidon defined by the vectors A, B, and C. Hence the box product of three vectors is zero iff (= if and only if) they are coplanar.
Proofs are obvious.
Theorem: For any three vectors A, B, and C, their triple cross-product is the vector
A x (B x C) = (AC) B - (AB) C
Stated without proof.
Corollary: This vector lies in the plane (B, C).
Proof is obvious.
Theorem: For a spherical triangle with vertices A, B, C on a unit sphere, the magnitudes of the cross-products of the vectors to the vertices are the sines of the included sides a, b, and c.
|| B x C || = sin a, || C x A || = sin b, and || A x B || = sin c.
The inner (dot) products are the cosines of the included sides.
BC = cos a, CA = cos b, and AB = cos c.
Corollary: (B x C)(C x A) = - (cos C / sin c) sin a sin b sin c
and its cyclical permutations. As a mnemonic aid, notice that, on the left-side, the vertices are in lexicographical order, with the second vector repeated; while, on the right-side, that vertex and the side-opposite appear in the fraction.
Proof: Hint the inner (dot) product is that of the exterior angle, while the vertical angle is interior -- supplementary.
Theorem: The sine of the angle between a point P on a sphere and a plane is the inner (dot) product of the vector P and the pole of the plane, divided by the product of their magnitudes.
Given a spherical triangle with the vertices A, B, C, on a sphere of radius rho. We employ the same symbols for the vectors from the origin to these vertices. The sides opposite are designated by the corresponding lower-case letters.
Draw a plane through these three points A, B, C. These same points are vertices of the corresponding plane triangle. Since each of the plane-congruence theorems carries over to the spherical case, the same constructions may be employed in the spherical case, as those of the plane case, to find the circum-center and the in-center. Indeed, the same vectors result.
The circum-center is at the intersection of the perpendicular bisectors of the sides of the triangle
We proceed by finding the poles of these bisectors. Make the replacement B --> A and A --> (A + B) in the orthogonalization theorem. Then the vector
U = A - (A + B)((A + B) A) / ((A + B)(A + B)) = (A - B) / 2
is orthogonal to A + B and in the (A, B) plane. Thus U is the pole of the perpendicular bisector of the side (A, B). Likewise, the replacement B -> A and A --> (A + C) yields
V = A - (A + C)((A + C) A) / ((A + C)(A + C)) = (A - C) / 2
which is orthogonal to A + C and in the (A, C) plane. This V is the pole of the perpendicular bisector of the side (A, C).
Their cross-product is the intersection of the two perpendicular bisectors.
4 U x V = B x C + C x A + A x B
is the circum-center of the spherical or plane triangle For future reference, its square is
(B x C + C x A + A x V)^2 =
sin^2 a + sin^2 b + sin^2 c - 2 (cos A / sin a + cos B / sin b + cos C / sin c) sin a sin b sin c
The cosine of the angle between this pseudo-vector and either of the vertices is
cosine(angle of circum-radius) = ABC / (rho ||B x C + C x A + A x B||)
The actual circum-radius is the sine of that angle; i.e., sqrt(1 - the square of the preceding expression). This formula provides the radius of the circum-circle, in the spherical case. For example, if A = i, B = j, and C = k; then the cosine of the angle of the circum-radius is sqrt(3) / 3. Thus, the circum-radius becomes sqrt(6) / 3.
The in-center is at the intersection of the bisectors of the angles of the vertices of the triangle.
We proceed by finding the poles of two of these bisectors
(A x B / ||A x B|| + A x C / ||A x C||) / 2
(B x C / ||B x C|| + B x A / ||B x A||) / 2
Their cross-product is the in-center of the triangle. Make the replacement a --> A x B, B --> B, C -->C, etc. in the theorem which gives the triple cross-product
(A x B) x (B x C) = ((A x B)C) B - (A x B)B) C = (ABC) B
etc. Their sum is
ABC (A sin a + B sin b + C sin c) / (4 sin a sin b sin c rho^2)
It is the inner (dot) product of the poles of the aforementioned bisectors. A collinear vector is
A sin a + B sin b + C sin c
For future reference, its square is
(A sin a + B sin b + C sin c)^2 =
sin^2 a + siin^2 b + sin^2 c +2 (cot a + cot b +cot c) sin a sin b sin c
The sine of the angle between this vector and the plane of a side is
ABC / (||A sin a + B sin b + C sin c|| rho^2)
This formula provides the radius of the in-circle, in the spherical case. We can proceed even further. Previously, we had provided formulae for the angles between pairs of sides, e.g., sin a = || B x C || / rho^2 . Substituting, we obtain
ABC / (|| (A || B x C || + B || C x A || + C || A x B ||) ||)
For example, if A = i, B = j, and C = k; then the in-radius is sqrt(3) / 3.
Copywrite © 1999 R. I. 'Scibor-Marchocki last modified on Sunday 30-th May 1999.