Study of space analytic geometry with the theory of vectors and matrices.
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English Pages [186] Year 1949
Table of contents :
COORDINATES AND LINES
PLANES
SURFACES AND CURVES
SPHERES
QUADRIC SURFACES
THEORY OF MATRICES
ROTATIONS OF AXES AND APPLICATIONS
SPHERICAL COORDINATES
ELEMENTS OF PROJECTIVE GEOMETRY
8]m > a: 73
OUP
88 15-8-74
1
5.000
OSMANIA UNIVERSITY LIBRARY Call
Accession No.
No. ^""16
Author
"^^ i &
\
A'M S
Title This bonk should be
re^rned
t
n or before the date
last
marked below.
In this
new
position
of
text the author presents an exanalytic geometry of threedimensional space. The material covers the standard topics of space analytic geometry but provides a treatment of the subject which permits immediate generalization to n dimensions. This treatment ties the subject to modern mathematics, and, in particular, to modern algebra. The use of the theory of vector spaces and matrices permits a major simplification in the proofs and in the exposition in general. Thus the aim of the book is to provide a modern and simpler treatment of the subject matter which permits easy generalization and fits the subject into its proper place in modern mathematics.
the
The
early part of the text is simplified by the use of the concepts of inner and scalar product. brief but adequate chapter on the theory of matrices provides a full, clear exposition of the principal axis transformation in the n-dimensional case. An additional feature is provided by the chapter on spherical coordinates where the mathematics of the approximations used in actual physical measurements of direction and distance in rotated coordinate systems is presented. This material is r,?/,) 2 + (z/j/,-) 2 - ^x^XfUi = (xy/ < j ^ HJ and must be nonnegative. The numbers P P, Q Q, and (P Q) 2 are all have shown that
an angle
6
is
expressions
The square x^yS + xfy?
the form
2o: t ?/ t x ; 7//
for
of all expressions
of the
It follows that there exists
+ yS} all
x/y
2 t
)
for
positive,
1
j
=
y'
+
2/0,
z
=
z'
+
ZQ,
sets of coordinates.
z9
FIG.
Formula axes, which
7.
sometimes used
(16) is
simplifies
what we
in
making a translation
shall later call
of
an equation of a
surface.
EXERCISES
What
1.
surfaces?
,. W (6)
-
(^
2
l)
+
-
2
(e) )
(z-
2)
3)*
_ ~
9~~
e
- 3) + (z + I) = 5 - 12 = x 2x + 3y 6?/ + 42 2 = 9 z + Go; 2x + 3y 12z/ + z 2 = 6a; 3a; 3?/ + Oz 9y + 24z 3x - 2/ 2 + s 2 = 6 2(x
2
2)
-
2
2
2
(y 2
2
(d)
+
(y
,
-I
4
2
(c)
translation of axes will simplify the following equations of
2
2
1
COORDINATES AND LINES
Sec. 12]
15
Find the
2.
there
is
#', y', z' coordinates of each of the following points a translation of axes with origin at 0'.
P
if
P=
(1, -1,1), 0'= (2,2, -1) = (b)P (0,0,0), 0'=(1, -1,1) = (-1, -1,1), 0'= (-1, -1,2) ( C)P (d)P = (3,1,1),0' = (-2,2,1) (e)P = (l,2, -3),0' = (1,0, -1) = (2,4,0),0' = (0, -1,0) (/) P
(a)
3.
Let the vectors
responding points. 4. Find the x',
P
in Exercise 2
Give their y', z'
x, y, z
be the
a/,
?/',
2'
coordinates of cor-
coordinates.
equation of the surfaces given by the equamoving the origin to
tions of Exercise 1 after a translation of axes
0'
=
(1,
12.
P*
=
-1,2).
Geometric addition of vectors. Let P\ (xi, y\, Zi) and Then Pi#P 2 determines a plane (#2, 2/2, 22) be two vectors.
FIG.
8.
and, as in Fig. 8, we can determine a point Pa in this plane such that OP 3 is parallel to PiP 2 and in a corresponding direction, |OP 3 = |PiP 2 |. The coordinates of P 2 relative to an x' y', z 1
,
,
|
coordinate system with origin at Pi, will be (x 2 z 2 ~ Zi). But P 3 is located with respect to the exactly where P 2 is located with respect to the x' y', ',
Xi,
y(Z2
be computed by the use of
+
[\(xi
(2/1
-
-
y2)
x2)
2
+
+
(zi
M(?/I
~
-
z>)*
2/2)
+
numbers are given rather than direction cosines, it is necessary to compute direction cosines before formula (29) can If direction
be applied. EXERCISES
Compute the
distances between points and lines in the following cases:
P,- (0,0,0);^ =
^=^
P,= (1,0,0);*"- =
V
(6)
(c)
P =
=
(a)
2
(l,
-1,0);-
~
-
=
7
~
=
= 62 + 12 2)/ = 2 7/- l=2e-4 (1,1, -1);2* (e) ~ g " 1 _ _ z -t( 11 IV ~ +J ~ ( r\P (J) * 2 1, i, i),
(d)
P2 = P =
(0, 1, 2);
3z
=
2
M w
P r 2
re 7 2V w, /, ^;,
x
+
2
3
-
^il
2 l
-
1-
2
CHAPTER PLANES
2
If we draw a ray from the origin O per1. The normal form. pendicular to a given plane, the ray is called the ray normal to the plane and the line of which it is a part the normal line. The will direction cosines have which are deterX, /z, v, ray uniquely
mined unless the plane passes through the case the selection of X,
ju,
v,
rather than
But
origin. X,
v,
ju,
in this
will
not
affect our results.
Let p be the distance from to the given plane so that p ^ Then the normal will intersect the plane in the point P
0.
A
translation of axes, which moves the origin to the coordinates x, y, z of the arbitrary point P by its replaces Po, z' coordinates #', y', (pX, pp., pv).
(x
-
y
pX,
-
*
p/z,
-
pv).
The ray through P and the vector whose transformed coordinates Then P is on the plane (X, /z, v) is normal to the given plane. if and only if the vector just defined is orthogonal to PoP, that is, are
if
and only
if (X, M,
v)
pX, y see that
(x
pp, z
pv)
=
0.
We com-
P is on the given plane if and pute this inner product and 2 2 2 ^ ) = 0. vz if \x Since X 2 + M 2 + M py p(X only + u 2 = 1, we have proved that P (#, y, z) is on the given plane if and only if
+
+
\x
(1)
We shall
call
formula
+
y
+
+
vz
=
p.
equation of a plane in normal form. p > 0. When p == 0, the equation
(1) the
completely unique when is equally valid and the selection of one set vz = M?/ of direction cosines rather than their negatives may be regarded as implying that a positive direction on the line normal to the plane has been selected. This is of little importance. It is
\x
EXERCISE
Give the equation in normal form of the two planes p units from the origin and normal to the line joining Pi to Pi in the following cases: 24
PLANES
Sec. 2]
P = P! = = 1, Px = iP = 7, P!
p = p = = (c) p = (d) p () p =
(a)
5,
(6)
(-1, (-2,
l
2,
2
-1),
1,
(0, 0, 0),
2
P2 =
az
(2)
+
(1, 3,
(2, 2, 1)
-1,2) -3, 8)
(0,
=
(6,
Every
equation.
by
-1) -2)
(1, 2,
P =
(2,l,4),P 2 = (4, -1, 6), P,
1
The general
2.
P -
2, 3),
25
+
cz
linear equation
+
=
d
0,
d are real numbers and a, b, c are not all zero, is an equation of a plane; for we can write this equation in an equivalent form
which
in
a, b, c,
tax
(3)
where
=
e
1
if
d
^
and t
(4)
+ e
if it
(2) is
(6)
=
ecz
1
if
d
>
\d\,
(3)
by
t
The number
0.
0,
to obtain
the equation of formula (2) if and But then formula the equation of formula (5). an equation of the plane for which
point
only
P
=
+
Va r+W+~c'- >
=
and we may divide formula
A
eby
(x, y, z) satisfies
satisfies
p
= M,
x-,
=
=
,
the plane p units from the origin and having a normal ray whose direction cosines are the numbers X, /z, v of formula (6).
i.e.,
It is
important to observe that the original coefficients a, 6, c numbers of the normal line. abed ? 0, the general equation may be written in the form
are a set of direction If
This equation
is
called the equation of a plane in intercept
and the numbers /o\ (8)
=
dj,d /--^ ,
g
are called the intercepts of the given plane.
=
form
~ d
They
are indeed the
SOLID ANALYTIC GEOMETRY
26
[Chap. 2
nonzero coordinates of the three intercept points (6,0,0),
(0,0,0)
(0,/,0),
where the given plane cuts the coordinates axes. If a plane defined by formula (2) passes through a point and Pi = (xi, y\, zi), then ax\ + byi + cz\ + d = a(x
(9)
-
xi)
+
-
b(y
y,}
+
-
c(z
zi)
=
0.
This equation involves the coordinates of PI and a set of direction numbers of the normal line. We shall call it the paint, direction number form of an equation of a plane. EXERCISES 1.
Use formula
(9) to
write an equation of a plane through Pi perP 2 to PS in each of the following cases:
pendicular to the line joining (a)
=
P!
(-1,
2, 3),
P = 2
(-3,
1, 2),
P =
(6) (c)
(d) (e)
(/) (g) 2.
(a) (6) (c)
(d) 3.
4.
P = P =
P =
(-5,
3
P =
(-1,
(1,2,
1
(2,
-1,4),P 2 = (-3,1, -2),P 3 =
= = P = P! =
(-1, (-2,
1
(1, 4,
Find
(3, 2,
p, X,
-
ju,
P = P 1, 3), = P -1), = P -2),
2,
-3),
2
2
v for
(4,
-1,
2),
3
+
20
=
0.
4,
P 4 = (9, 10, 12) -8, 0), P 4 = (6, -8,
2
(6, 7, 8),
2
(-9,
7)
each of the following planes:
+z = 6 3x 6y + 7z = -1 3 + 2y + 2z = -6 -3z + 6y + 22 = 14 2x
4z
-1) -4) = (-1, 4, 3), P 3 = (0, -3, -4) = (3, -4, 5) (4, -6, 7), P 3
(1,
-2, -1),
Pa P,
-
3y
l
2
4, 6)
2x
Ans.
2y
Find the intercepts of each Find a set of values of X,
(e)
(/) (g)
(h)
+ 2x - 3z = 28 + 4y + 62 = -49 - 2z = 6 x + -x + y + 2z = -3 6y 12x
T/
of the planes in Exercise 2. v for each of the following planes /x,
through the origin:
(b)
+y= 3z + 4y =
(c)
3z
(a)
x
-
4?/
=
(d) (e)
62
(/)
+ y = 2z -x + y + 2z = 3o: + &y + 2z =
2x
Write an equation of each of two planes parallel to a corresponding plane of Exercise 4 and one unit from the origin. 6. Write a set of equations in symmetric form of a line perpendicular to the given plane and through the point ( 1, 2, 3) in each of the cases 5.
of Exercise 2.
PLANES
Sec. 4]
3,
=
Planes
three
through
and P =
27
=
Let Pi
points.
P
(xi, y\, Zi),
2
be three distinct points. By Sec. 9 of Chap. 1 the three points are collinear if and only if a scalar multiple of (x 2 is (x* yi, 23 xi, 2/3 Zi) #j, (#2, yzj zz),
2/2
22
2/1,
If
2/3, z-6 )
(#3,
3
21).
a plane passes through Pi,
b(y
+
2/0
P we 3,
c(z
=
Zi)
its
equation
If it also
0.
is
a(x
+
#1)
passes through
P
and
2
have the relations
(10)
a(z 3
-
+
a?i)
-
6(2/3
+
2/1)
c(z 3
~
zi)
=
0.
do not have proportional coeffiare not Pi, collinear, and they can be solved of direction numbers (a, 6, c) = (0, 0, 0) of the to yield a set normal to the given plane. Then formula (9) is an equation of These two equations
in a, 6, c
P P
when
cients
2,
3
the required plane.
EXAMPLE
ILLUSTRATIVE an
Find
the
of
equation
plane through
1,2,3),
(
3,1,2),
(
(-5,4,6). Solution
Equations (10) become
-2a 4a
Then -46 = a(z
+
1)
+
oc,
6(y
8a
-
=
2)
c,
+
-
=
and
x
-
10t/
+
82 4-
1
26
so that
c(z
+
20
b
-
-
c
3c
-
-46 = 40,
3)
a(x
+
1)
-
24
=
x
equation of the plane.
= -10a.
b
10a(y
-
IQy Ans.
-
+
Then
2)
+
Sa(z
Sz
-
3
=
IQy
+
x
-
3)
= an
is
82
=
3.
EXERCISE Find an equation of the plane through Pi,
P
2,
PS
for each set of points
listed in Exercise 1 of Sec. 2.
Parallel planes have the same normal line. 4. Parallel planes. Then every plane parallel to ax + by + cz + d = is obtained if we leave a, &, c fixed and vary d. If we put an equation of a plane in normal form \x + ny + vz = p, we obtain all planes parallel to the given plane and on the same side of the origin by varying p ^ 0. The distance
between the given plane and a second plane \x
+
Mj/
+
vz
=
p\
SOLID ANALYTIC GEOMETRY
28
then clearly
is
\p
p\\.
[Chap.
however, we consider
If,
all
2
planes
parallel to the given plane and on the other side of the origin, we see that their equations are (\x ny vz) p lw Then the
distance between Xx is
p
+
+
ny
+
vz
=
+
+
= p and
(\x
+
+
ny
=
vz)
p\
pi.
ILLUSTRATIVE EXAMPLES I.
+
4z
Find an equation of a plane which = 7 and which contains the point
is
parallel to the plane 3#
P =
(
2y
2, 3, 3).
Solution
The equation may be taken
to have the form 3x
P
2y
+
4z
=
k
y
on the plane. Then 6 6 + 12 = = k and the required equation is 3x 2y + 4z = 0. II. Find the distance between the plane 3x 2y + 4z = 7 and the = 4z 12. 3;r + 2y plane
where k
is
to be determined so that
is
Solution
The normal forms of these two planes are obtained by multiplying equations by (9 + 4 + 16)~* = (29)""*. The distance is then
the
(29)-*12
-
=
5(29)"*. (29)"*7 III. Find the distance between the plane 3x 15. 82 = the plane 60; 4?/
2y
+
4z
=
7
and
+
Solution
distance is 7(29-*) Y(29)"* = ^(29)-* = \ \/29. IV. Find the equations in normal form of the two planes that are
+
The
five units
from the plane
x
+
2y
-{
2z
+
3
=
and
parallel to
it.
Solution
The normal form
of an equation of the given plane is obtained by 4 the (1 J and thus is 4)~* = given equation by multiplying
+ +
x-2y -
2z
3
4 units from the origin where we planes are then 6 units and as equivalent to the replacement of X, ju, v interpret a negative distance 2z = 18 for the v. The answers are then x 2y n, by X,
The two
value 6 and x
2y
2z
=
12 for the value
4.
EXERCISES 1. Find an equation 6f a plane parallel to the given plane and through the given point in the following cases:
PLANES
Sec. 5]
(a) (6) (c)
(d) (e)
(/) (g)
3x 2x
-
2y
+z=
1,
89
(-1, -2,1)
+ 2y-z = 0, (-1,1,1) z + + 3z = 4, (-2,1,1) 2x-y-2z = 7, (3, -1,3) 3* + 4y - 62 = 183, (4, 5, 2) I8x - Uy + I7z = 987, (1, 1, 0) Ite - 7y + 6z = 111,968, (2, 3, 1) 2/
Find an equation of a plane parallel to the given plane and having
2.
x-intercept 3 in each of the cases of Exercise 1. 3. Find an equation of a plane parallel to the given plane and two units further from the origin in each case of Exercise 1 except (e), (/),
and
(0).
Find the distances between the following pairs of planes: 2x + 2y - z = 3, 2x + 2y - z = 18 - y - 2z = 6, -2x + y + 2z = 12 (6) 2x
4.
(a)
(c)
(d) (e)
3x 6x
+
2y
+
6z
=
14,
3z
+
2y
+
6z
=
21
- 2y + 32 = 28, 12x - 4y + 62 = 35 18x - 6y - 9z = 35, -Go; + 2y + 82 = 21
Find equations of pairs of planes four units from the
5.
first
of the
planes in Exercise 4.
Distance from a plane to a point. If an equation ax + by d = of a plane is given and a point P = (zi, yi, z\) is cz given, we require a formula for the distance d from the plane to P. We first observe that we may divide the equation of the plane by c 2 to convert it to the normal form 62 \/a 2 5.
+
+
+
+
\x
Then the equation
+
ny
is
on this
8
(11)
>
If 5
0,
if 8
=
Xzi
+
is
\x
M2/i
+
+
P and the origin < 0, the point P and
the point
and
=
vz
plane, side of the plane.
p.
of the plane parallel to tue given plane
from the origin plane, we have
units further
P
+
py
*i
+
~
vz
= p
+
5.
and
d
Since
P-
are on opposite sides of the the origin are on the same
the given plane passes through the origin, the meaning of the sign of 5 is not clear, and we shall simply use the formula If
ILLUSTRATIVE EXAMPLES Find the distance from the plane 2x -1,5).
I.
(2,
+
2y
z
=
3 to the point
SOLID ANALYTIC GEOMETRY
30
[Chap. 2
SoliUion
The
distance
is
+
[2(2)
Find the distance from 2x
II.
-
2(-l)
+
(5)
2y
z
3]
= -2.
=
to
1, 5).
(2,
Solution
The
distance
is
|i(4
-
-
2
=
5)|
1.
EXERCISES
Find the distances from the following planes to the corresponding
1.
points:
(d)
+ 2y - z = 3, (5, 3, -2) + y + 2z = -2, (2, -2) 3z + 2y + 6z = 14, (1, -2, 1) 12x-4y + 6z = 35, (-1, -1,1)
(e)
18*
(/)
-6*
(a) (6) (c)
2x
-2*
1,
-
9y 2
-
+
Find the the plane 3x 2.
+
(-1,
2,
-3)
+
Solve Exercise 2
3.
(1, 1, 3)
coordinate of a point P = Qz = 7 to the point 2y
z
+
= 14, 3z = 0,
9z
if
the distance
is
3, 2, 2) if
(
P
the distance from
is 2.
4.
Angle between two planes. The angle 6 between two planes normal to the defined to be the angle between the rays from 6.
is
two
planes.
It is not
through the origin. Xix
+ my +
v\z
=
uniquely defined
if
either plane passes
+ p,y + vv\.
the equations are \x = XXi MMi pi, then cos If
+
equations are given in general form, they into normal form.
must
-f~
first
= p and When the
vz
be converted
EXERCISES 1.
Compute the
planes:
+ 2y - z = 3,3x - 2y + 6z = + y + z = 2x - y - 2z = 6 - y + z = 2, 2x + y = 4 + = 3, 3z + 4y = 2
(a)
2x
(6)
x
(c)
x
(d) 2.
(a) (6) (c)
7.
a?
cosine of the angle between the following pairs of
Show
T/
that the following pairs of planes are perpendicular.
+ 2y - z = 3, x - 2y - 2z = + y + z = 1, x - z = 6 3x - 27/ + 4^ = 1, 2x + y - z =
2x *
The
14
1,
line.
If
1
two planes. Two nonparallel the planes have as equations
line of intersection of
planes intersect in a
1
PLANES
Sec. 7]
+
+ +
biy b zy
31
+ c z +
ciz
di
2
d*
= =
0,
then their line of intersection is orthogonal to both of their normals and the direction cosines of the line satisfy the equations
a 2X
+ +
+ &2M + bin
= =
civ c 2 j>
0.
However, the simplest method of determining the line two points on it. The procedure is illustrated below. ILLUSTRATIVE
is
to find
EXAMPLE
Give a set of equations in symmetric form of the
line
L
of intersection
of the planes
2x
+y-
=
22
5x
1,
and determine direction cosines
+
4y
-
6z
82
=
1
=
= ~2
of L.
Solution
We solve for x and y by writing 8x + =
ty
4 therefore 3x ;
6 and
x
Put Put
2 2
= =
0, 3,
=
2s
+
6 >
t/
^
=
2z
2^
+
22
- 9 -g
and obtain P t = (2, -3, 0) on L. Then L has as and obtain P 2 = (4, 1, 3) on L.
-
x
2
_ "
y
+
3
its
equation
_ ~
2
2
2z
3
223
and the corresponding direction cosines are
Vl7'
VT7' EXERCISE
Give a set of equations in symmetric form of the line intersection of the following two planes, and determine its direction cosines, - 2y z = 5, x y z = 1 (a) 3z (ft)
(c)
(d) (e)
(/) (g)
+ + + + +
z = s - 2y 2z Zx 4y - 3s 6s 2y 4x - j/ + 2 = * - V - 32 = - 2z 4x 3y
+
+ +
6,
= =
re
1,
-
z
x
-
=
2
y
+
z
=
6
-y+2 = 6 -5, x - 3^ = 4 2, 2x + y - 2 = -3 2, 2x + 2y = x + 2y - 3z = 4 1
,
SOLID ANALYTIC GEOMETRY
3* 8.
+
cz
Angle between a line and a plane. + d = and the line
x-
xi
(14)
_
-
y
ai
=
yi
[Chap. 2
Let the plane ax
s
bi
-
+
by
z\
ci
between the line and the be given. We require the angle This angle is defined to be the complement of the angle plane.
between the directed normal to the plane and the line and is not unique unless the line is also directed. When both are directed, we will have 6
where we have now made
=
cos
and
X, M> v
Xi, pi, v\
unique.
Then
and
sin
sin
(16)
=
XXi
+
MMI
+
vv\.
EXERCISE Find the sine of the angle between the following lines and planes: - 2y + 4z = 1, 3x + 4 = 2y - 1 = 6z + 7 (a) 3z
+
(6)
x
(c)
3z
2y
+
2y
Pencils
9.
+
biy define
+
CiZ
two
-
=
2z
-
62
of
+
=
di
=
4,
planes.
Let
and g(x,
distinct planes.
(17) is
=
=
=
2,
sf(x, y, 2)
= the
y, z)
2-3 equations f(x, yS$) = a\x = a 2x b*y c& d* =
+
+
+
Then the equation
+
tg(x, y, 2)
=
called the equation of the pencil of planes determined
by the
two given planes. Indeed, formula (17) is an equation of a plane for all real numbers s and t which are such that the vector When the two given planes are s(a,i, 61, Ci) + t(a 2 6 2 02) 5^ 0. ,
,
not parallel, they intersect in a line L; formula (17) represents a plane Q(s,
41
Thus we do not include planes
I.
in the set of
surfaces called cones.
A polynomial f(x,
to be homogeneous
y, z) is said
terms of
if all
have the same degree n. A homogeneous polynomial of is called an nic form. We shall study quadratic forms, forms of degree two, later. If f(x, y, z) is homogeneous of
f(x, y, z)
degree n i.e.,
degree n, then
n
>
7
f (x
+
x', y', z'
(x
1.
n
f(x, y, z).
Also /(O,
0, 0)
=
+
XQ, y'
then
,
,
?/o,
if
yo, z'
+
=
f (x, y, z)
y', z')
homogeneous of degree n > 1 in an equation of a cone with vertex at
ZQ) is
is
z
zo),
which moves the origin to
of axes,
replaces /(z, y, z) by g(x' y', z'}. is on the surface S defined
Since g(Q,
,
V
the point
Q =
has the property that g(x',
f (x, y, z)
//
). y For a translation
(XQ,
t
0.
Theorem
=
=
f(tx, ty, tz)
7
(xi, yi
=
=
f(x, y, z)
If
0.
and
a point on S, then g(xi, yi, Zi)
is
zi)
,
by
0, 0)
V = = 0,
for every t. But t(xi', */) = 7 7 = z of of is form a set the vector ) parametric (x y VL, Zi) equations of the line joining V to Q, and we have proved that all points of this line are on S, S is a cone.
g(txi', tyi', fe/)
t
n
7
g(xi',
?/i
,
1
,
',
The
cones, which consist of all points on the lines joining the vertex to the points of a fixed plane curve, are of particular Let us then derive the equations of such interest in geometry.
surfaces in the case where the given curve
plane and (XQ,
?/o,
cone
if
is
ZQ)
equation the vertex.
and only
if
P
a point on the curve. (2)
X
-
XQ
=
-
t(Xi
a curve in the
is
is
(x,
Z Q 7* 0.
ZQ(X
P
assume that
y
XQ),
-
y)
7/o
= %l ~
2/o),
is
ZQ
tZQ.
not a point of the x, y plane and thus t and obtain
Then we may eliminate XQ)
=
-
(Xi
XQ)(ZQ
-
z)
,
Z Q (y
y
)
=
(j/1
~
J/o)
(ZQ
We /ox
(3)
y
0.
Z
We
x,
Suppose that PO = Then a point P = (x, y, z) is on the P = t(P\ PO), where Pi = (x\, y\, 0) is Consequently
its
solve for Xi
xi
=
and
^ H x
,
y\ to obtain
ZQ(X
-
XQ)
ZQ
Z
=
z Q (y
__ 9
7/1
t/o
-
y
-\
ZQ
Z
)
-;
^
Z).
SOLID ANALYTIC GEOMETRY
42
[Chap. 3
and substitute these formulas in (xi, y^ = 0. The resulting equation in x y, z is an equation of the given cone. y
ILLUSTRATIVE EXAMPLES I. Find an equation of the elliptic cone with vertex at 2 base the curve 4# 2 y = 1.
1, 3),
(2,
and
+
Solution
We use
formula
(3)
and write
-
3(x
Then
equation
=
z)zi
(3
3#
-
2)
(3
2z,
0)1/1
=
3y
+
3(y
+
z,
1)
so that the required
is
4(3*
-
22)
2
+
(3ff
+
2
2)
=
(3
-
2
2)
.
The
solution should be left in this form since the equation is not essentially simplified when the indicated operations are carried out.
Find an equation of the hyperbolic cone through the point (3, 0, 0) 2 2 = 1 the and (?/ /4) hyperbola whose equations are z = 0, (x /9) II.
.
Solution
y
We use the method of derivation of formula (3) to write x - 8)2,, = ty,, z = tzi. Then -3 1. one line of S through each point of S. :
For if P is any point of and therefore assume that
+ /n -i(s, y) + of degree k.
x
=
ta,
0(0
=
y
/S,
we may
/(O, 0)
=
translate the origin to
Then
0.
f(x, y)
tft,
z
line L through the origin P = ty, and = */n(, 0) + f^Vn-iCa,
P y)
has equations
/?)++
0)
/(te,
f
n (x,
+ fi(x, y) where each/fc(, y) is homogeneous
-
Every
=
=
1 and that f(x, y) is and the only line through P is irreducible. Hence, a = ft = the z axis. EXERCISE that
ftx
f(x, y)
Show that a surface f(x, y) = determined by a homogeneous polynomial f(x, y) of degree n consists of r ^ n planes through the z axis or is a point surface. 11.
+ -Pi is
ty,
=
Tangent z
=
+
Zo
y = yQ in a point
intersect a surface /(x, y, 2) real number
=
if t\ is
+
x
XQ
=
+ = ^o + ,
=
T/O
ft?,
2o
t%,
+
ti
which