Calculus: Analytic Geometry and Calculus, with Vectors

110 Vectors and geometry in three dimensions

matrix. Therefore U'1 = UT. This is important; the inverse of a unitary

matrix U is U.

19 This problem requires us to agree with Miss Garnett that methods of

analytic geometry can be used to solve a challenging problem that may baffle

those who seek more elementary solutions. It is supposed that a, b, c are positive

numbers and that the points 4(0,0), B(c,O), C(a,b), D(a + c, b) are vertices of a

parallelogram. Let a point E(u,O) on the bottom side of the parallelogram be

joined to the top vertices C and D and let a point F(t,b) on the top side of the

parallelogram be joined to the bottom vertices A and B. The lines EC and F11

intersect at P,, and the lines ED and FB intersect at P2. The line P1P2 meets

the side SIC at Q, and meets the side BD at Q2. The question then arises whether

the distance from Q, to A is equal to the distance from Q2 to D. Elementary

geometrical considerations show that the answer will be affirmative if the line

P,P2 contains the center P3 of the parallelogram, this center P3 being the inter-

section of the diagonals of the parallelogram. Show that, for each k = 1, 2, 3,

the elements of the kth row of the determinant

to bu

t+u-a t+u - a

ac + cz to b(c - u)

1

a+2c-t -u a+2c-t -u

a+c b

2 2

are xk, yk, 1, where xx and yk are the coordinates of P. Then prove that the

determinant is 0 and hence that the line P1P2 actually does contain P3.

20 Let P1(x,,y,), P2(x2,y2), P3(x3,ya) be vertices of a triangle such that no two

of the vertices lie on a line through the origin. Let X1, Xs, X3 be three different

numbers, and for each k = 1, 2, 3, let Qk be the point Xkyk) The two tri-

angles P1P2P3 and Q,QtQ3 are then perspective, the center of perspectivity being the

origin. The lines P1P2 and Q,Q2 intersect at a point R3, the lines P2Pa and Q2Q3

intersect at a point R,, and the lines P3P1 and Q3Q1 intersect at a point R2. The

famous Desargues theorem says that the three points R,, R2, R3 lie on a line L. It

is easy to sketch figures illustrating the theorem, but proofs are not easily origi-

nated. Possessors of sufficient time, paper, and technique may cultivate addi-

tional technique by finding the x coordinate of R3, and then interchange x and y

and advance subscripts to discover that the coordinates of R2 and Ra are the first

two elements of the bottom rows of the determinant in the equation

x y 1

(X2 - 1)X,x, - (X, - 1)X3x3 (X3 - 1)X,y, - (X1 - 1)X3y3 1

x3 - x, x3 - X,

(X1 - 1)4V2 - (X2 - 1)X,x, (X, - 1)a2y2 - (A2 - 1)11x1

X, - 1\2 X1 - Xz

=0.

This is, therefore, the equation of the line L through Rz and R3. Considerable

courage is required to show that the coordinates of R, satisfy this equation and

thus obtain an analytic proof of the Desargues theorem.