Mathematical Tools for Physics

11—Numerical Analysis 339

y

x

Do this in two dimensions, fitting the given data to a straight line, and to describe the line I’ll use vector
notation, where the line is~u+α~vand the parameterαvaries over the reals. First I need to answer the simple
question: what is the distance from a point to a line? The perpendicular distance from ~wto this line requires
that
d^2 =

(

~w−~u−α~v

) 2

be a minimum. Differentiate this with respect toαand you have

(~w−~u−α~v

)

.

(

−~v

)

= 0 implying αv^2 =

(

~w−~u

)

.~v

For this value ofαwhat isd^2?

d^2 =

(

~w−~u

) 2

+α^2 v^2 − 2 α~v.

(

~w−~u

)

=

(

~w−~u

) 2

−

1

v^2

[

(~w−~u).~v

] 2 (49)

Is this plausible? (1) It’s independent of the size of~v, depending on its direction only. (2) It depends on only the
differencevector between ~wand~u, not on any other aspect of the vectors. (3) If I add any multiple of~vto~u,
the result is unchanged. See problem 37. Also, can you find an easier way to get the result? Perhaps one that
simply requires some geometric insight?
The data that I’m trying to fit will be described by a set of vectors~wi, and the sum of the distances squared
to the line is

D^2 =

∑N

1

(

~wi−~u

) 2

−

∑N

1

1

v^2

[

(~wi−~u).~v

] 2