Math 5335 Section 2
Fall Semester 2005
Selected solutions: 2nd HW assignment
§1.4, Problem 28. The point Q(5,-2)
is the vertex of the angle. We call the given points on the two sides of
the angle P = (4,-1) and R = (0,0).
The line joining P and Q is given
in normal form as 
(1,1),X
= 3, or equivalently
x + y = 3. The line joining
Q and R is given in normal form as
(2,5),X = 0,
or equivalently 2x + 5y = 0.
By definition, the interior of the angle consists of all points
that are on the same side of PQ as R and also
on the same side of QR as P. We
calculate: (1,1),R = (1,1),(0,0) = 0 < 3. This shows that the points on the same side
of PQ as R are characterized by the inequality
(1,1),X < 3, or
equivalently x + y < 3.
A similar calculation shows that
(2,5),P = (2,5),(4,-1) = 3 > 0.
This shows
that the points on the same side of QR as
P are characterized by the inequality
(2,5),X > 0, or equivalently
2x + 5y > 0. Therefore, the interior
of the angle is characterized by the following system of inequalities:
x + y < 3 ;
2x + 5y > 0 .
§1.4, Problems 72 and 76.
Let the lines be given parametrically as:
l: t --> P + tU
m: t --> Q + tU.
(We can use the same direction vectors for both lines, since they are
parallel.) Now, let Y = Q + tU
be a point on m, and let d
be the distance from Y to l.
According to Theorem 45 (more specifically, equation 1.21), we have:
| d² =
||Y-P||² - |
|
Y-P,U² |
| ------------ |
| ||U||² |
|
Making the substitution Y = Q + tU,
we obtain:
| d² =
||(Q-P)+tU||² - |
|
(Q-P)+tU,U² |
| -------------------- |
| ||U||² |
|
Replacing each square of a norm by the inner product of the vector with itself,
we obtain the following version of the formula:
| d² =
(Q-P)+tU,
(Q-P)+tU - |
|
(Q-P)+tU,U² |
| -------------------- |
| U,U |
|
Now, it this may not seem to look like progress as yet. But it has the advantage that the inner products can be expanded, and we'll then be able to simplify the
resulting formula considerably. Thus:
| d² =
(Q-P),(Q-P) +
2t(Q-P),U + t²U,U - |
| ((Q-P),U + tU,U)² |
| -------------------------------- |
| U,U |
|
Before we can cancel anything, we have to expand the numerator of the fraction.
This is not as bad as it might look, as the inner products that appear inside
the parentheses are ¡merely real numbers! Accordingly:
| d² =
(Q-P),(Q-P) +
2t(Q-P),U + t²U,U - |
| (Q-P),U²
|
| ----------------- |
| U,U |
|
- |
| 2t(Q-P),UU,U |
| ------------------------- |
| U,U |
|
- |
| t²U,U² |
| -------------- |
| U,U |
|
Cancelling the appropriate factors from the numerators and denominators of the
last two fractions, we obtain an expression in which the terms that involve
either t or t² cancel out. So, we do
all of that cancellation. And we also replace each instance of the inner
product of a vector with itself by the square of the norm. After doing
both of those things, we end up with the following expression:
| d² =
||Q-P||² - |
| Q-P,U² |
| ------------ |
| ||U||² |
|
Well, this looks like the formula that we started with, except
that Q is substituted for Y. And even
of more significance is the fact that the answer to this calculation
turns out not depend on the choice of a point on m,
so that:
- It proves what was claimed in Problem 72, namely that the distance
to l from any point
on m is the same as the distance
to l from any other point
on m.
- It provides the formula requested in Problem 76.
Finally, to finish Problem 76, you have to substitute the given data into the
formula. The answer turns out to be 0. But then you can check your
answer by verifying that the two lines actually are the same line.
§1.4, Problem 72: alternate solution. Let the equation of
l be given in normal form as A,X = c,
and let the equation of k be A,X = b. (We can use the same
vector A in both equations, since the lines are parallel
to each other.) According the formula in Theorem 23, the distance
to l from a point Y
is:
| |c - A,Y | |
| ____________ |
| ||A|| |
But if Y lies on k,
then A,Y = b. In this case, the formula for the distance
can be simplified to |c - b| / ||A||. This does
not depend on the choice of a point on k.
§2.5, Problem 11.
A solution will be written soon.
§2.5, Problem 15. Actually, the
problem stated in the text is erroneous, since both sides are the same ray.
So, instead of presenting a solution to the given problem (which would lead
to an angular measure = 0), I'll present the solution of the "old"
version, which called for finding the angular measure of 
(3,4)(2,-1)(4,5).
One side of the angle has direction indicator
(3,4) - (2,-1) = (1,5). The length of this vector is
261/2 [square root of 26 . . .],
so that we obtain the unit direction indicator
U = 1/261/2
(1,5). The other side of the
angle has direction indicator
(4,5) - (2,-1) = (2,6), which has
length = 401/2 = 2·101/2.
Therefore, the unit direction indicator of this side is
V = 1/401/2(2,6), or
V = 1/101/2(1,3). To find the angular
measure, we first calculate the inner product:
| U,V = |
|
· |
|
(1,5),(1,3) =
| |
(1,5),(1,3) =
| |
= |
| 8 |
 |
|
The angular measure is then given as the following integral:
 |
| 1 |
| |
| 8/ |
|
|
|
Back to the class homepage.
Prof. Joel Roberts
School of Mathematics
University of Minnesota
Minneapolis, MN 55455
USA
Office: 351 Vincent Hall
Phone: (612) 625-1076
Dept. FAX: (612) 626-2017
e-mail:
roberts@math.umn.edu
http://www.math.umn.edu/~roberts