Calculus: Analytic Geometry and Calculus, with Vectors

2.3 Scalar products, direction cosines, and lines in E, 91

quite mysterious until it is realized that the lines P1P2 and P3P4 might be hori-
zontal, while P'P" might be a vertical line perpendicular to both of them. We
shall solve this problem with the aid of vectors. The first step is to observe
that the coordinates of P' and P" can be written down easily if we find constants
X (lambda) and μ (mu) such that P1P' = APIPZ and F P" = uPsP4. Therefore,
our real problem is to find X and μ. Starting with the formula

P'P" = P'P1 + P1P3 + P3P"

P'P" = -APIA + !.L + PIP;.

The requirement that P'P" be perpendicular to PIP and to P3P4 is equivalent to
the requirement that 0 and PP"P3P4 = 0. This is equivalent to
the requirement that A and μ satisfy the equations

(3)

APIP2P1P2- μPP4'P1P2 = PP P P

AP1P2 P3P4- μP3P4P3P4= PIP. P.P4.

The question whether these equations have solutions for A and μ is now critical.

Such questions are so often critical that Theorem 2.57 will soon appear. For

the record, and perhaps for future reference, it can be noted that the determinant

of the coefficients of A and tc is

(4) - IP1P2I2 IP3P412 sine 0,

where 0 is the angle between the lines PIP2 and P3P4. If these lines are not paral-

lel, then the determinant is different from 0 and the equations uniquely deter-

mine A and μ. Our skew lines are not parallel. Therefore, A and s and hence

P' and P" are uniquely determined.

26 This problem requires that we pick up the idea that vectors and scalar

products are not unrelated to problems in statistics. Let n be a positive integer

and suppose at first that is = 3. Suppose n students took examinations in

English and Mathematics and got grades e1, e2, , e, and m1, m2, , ,

m,,. Let the mean (or average) of the English grades be B and let the mean of

the Mathematics grades be M. For each k = 1, 2, , n let

(1) uk=ek - E, Vk=Ink - M.

The numbers u1, n2, ,u,, and v1, V2, .. ,vn can be regarded as scalar

components of vectors u and v in Euclid space E. of n dimensions. Show that,

except in the trivial case in which Jul = 0 or JvI = 0, the cosine of the angle

between these vectors is determined by the formula

(2) cos 0 =

u1V1 + U 2V2 + + unyn

Iul IVl 1/u2l +u2 +. .. + un 1/vi + v2 + ... +

vn

Remark: It is not difficult to develop enough geometry of En to show that (2)

is valid when n > 3 as well as when n = 3. In statistics, the last member of

(2) is called the correlation coefficient of the English and Mathematics grades.

Show that if this coefficient is 1, then the vectors u and v must have the same