Handbook of mathematics for engineers and scienteists part 23 pptx

122 ANALYTIC GEOMETRY
3. Vectors a and b are collinear if and only if a × b = 0. In particular, a × a = 0 and
a(a × b)=b(a × b)=0.
4. (λa) × b = a × (λb)=λ(a × b) (associativity with respect to a scalar factor).
5. The cross product of basis vectors is
i × i = j × j = k × k = 0, i × j = k, j × k = i, k × i = j.
6. If the vectors are given by their coordinates a =(a
x
, a
y
, a
z
)andb =(b
x
, b
y
, b
z
), then
a × b =





ijk
a
x
a
y
a

z
b
x
b
y
b
z





=(a
y
b
z
– a
z
b
y
)i +(a
z
b
x
– a
x
b
z
)j +(a
x

b
y
– a
y
b
x
)k.(4.5.3.6)
7. The area of the parallelogram spanned by vectors a and b is equal to
S = |a × b| =




a
y
a
z
b
y
b
z



2
+



a

x
a
z
b
x
b
z



2
+



a
x
a
y
b
x
b
y



2
.(4.5.3.7)
8. The area of the triangle spanned by vectors a and b is equal to
S =

1
2
|a × b| =
1
2




a
y
a
z
b
y
b
z



2
+



a
x
a
z
b

x
b
z



2
+



a
x
a
y
b
x
b
y



2
.(4.5.3.8)
Example 1. The moment with respect to the point O of a force F applied at a point M is the cross product
of the position vector
−−→
OM by the force F; i.e., M =
−−→
OM × F.

4.5.3-3. Conditions for vectors to be parallel or perpendicular.
A vector a is collinear to a vector b if
b = λa or a × b = 0.(4.5.3.9)
A vector a is perpendicular to a vector b if
a ⋅ b = 0.(4.5.3.10)
Remark. In general, the condition a ⋅ b = 0 implies that the vectors a and b are perpendicular or one of
them is the zero vector. The zero vector can be viewed to be perpendicular to any other vector.
4.5.3-4. Triple cross product.
The triple cross product of vectors a, b,andc is defined as the vector
d = a × (b × c). (4.5.3.11)
The triple cross product is coplanar to the vectors b and c; it can be expressed via b and c
as follows:
a × (b × c)=b ⋅ (a ⋅ c)–c ⋅ (a ⋅ b). (4.5.3.12)
4.5. COORDINATES,VECTORS,CURVES, AND SURFACES IN SPACE 123
4.5.3-5. Scalar triple product of three vectors.
The scalar triple product of vectors a, b,andc is defined as the scalar product of a by the
cross product of b and c:
[abc]=a ⋅ (b × c). (4.5.3.13)
Remark. The scalar triple product of three vectors a, b,andc is also denoted by abc.
Properties of scalar triple product:
1. [abc]=[bca]=[cab]=–[bac]=–[cba]=–[acb].
2. [(a + b)cd]=[acd]+[bcd] (distributivity with respect to addition of vectors). This
property holds for any number of summands.
3. [λabc]=λ[abc] (associativity with respect to a scalar factor).
4. If the vectors are given by their coordinates a =(a
x
, a
y
, a
z

), b =(b
x
, b
y
, b
z
), and
c =(c
x
, c
y
, c
z
), then
[abc]=





a
x
a
y
a
z
b
x
b
y

b
z
c
x
c
y
c
z





.(4.5.3.14)
5. The scalar triple product [abc] is equal to the volume V of the parallelepiped spanned by
the vectors a, b,andc taken with the sign + if the vectors a, b,andc form a right-handed
trihedral and the sign – if the vectors form a left-handed trihedral,
[abc]=
V .(4.5.3.15)
6. Three nonzero vectors a, b,andc are coplanar if and only if [abc]=0. In this case, the
vectors a, b,andc are linearly dependent; they satisfy a relation of the form c = αa+βb.
4.5.4. Curves and Surfaces in Space
4.5.4-1. Methods for defining curves.
A continuous curve in three-dimensional space is the set of points whose coordinates satisfy
a system of parametric equations (4.5.2.19). This method for defining a curve is referred
to as parametric. The curve can also be defined by an equivalent system of equations
(4.5.2.18), i.e., described as the intersection of two surfaces (see Paragraph 4.5.4-2).
Remark 1. A curve may have more than one branch.
Remark 2. One can obtain the equation of the projection of the curve (4.5.2.19) onto the plane OXY by
eliminating the variable x from equations (4.5.2.19)

4.5.4-2. Methods for defining surfaces.
A continuous surface in three-dimensional space is the set of points whose coordinates
satisfy a system of parametric equations (4.5.2.17). This method for defining a surface is
referred to as parametric. The surface can also be determined by an equation (4.5.2.16) or
z = f (x, y).
Remark 1. A surface may have more than one sheet.
Remark 2. The surfaces determined by the equations
F (x, y, z)=0 and λF (x, y, z)=0
coincide provided that the constant λ is nonzero.
124 ANALYTIC GEOMETRY
Remark 3. For any real number λ, the equation
F
1
(x, y, z)+λF
2
(x, y, z)=0
describes a surface passing through the line of intersection of the surfaces (4.5.2.18), provided that this line
exists.
Remark 4. The equation
F
1
(x, y, z) ⋅ F
2
(x, y, z)=0
describes the surface that is formed by points of both surfaces in (4.5.2.18) and does not contain any other
points.
4.6. Line and Plane in Space
4.6.1. Plane in Space
4.6.1-1. Equation of plane passing through point M
0

and perpendicular to vector N.
A plane is a first-order algebraic surface. In a Cartesian coordinate system, a plane is given
by a first-order equation.
The equation of the plane passing through a point M
0
(x
0
, y
0
, z
0
) and perpendicularly to
a vector N =(A, B, C)hastheform
A(x – x
0
)+B(y – y
0
)+C(z – z
0
)=0,or(r – r
0
) ⋅ N = 0,(4.6.1.1)
where r and r
0
are the position vectors of the point M(x, y, z)andM
0
(x
0
, y
0

, z
0
), re-
spectively (see Fig. 4.34). The vector N is called a normal vector. Its direction cosines
are
cos α =
A
√
A
2
+ B
2
+ C
2
,cosβ =
B
√
A
2
+ B
2
+ C
2
,cosγ =
C
√
A
2
+ B
2

+ C
2
.(4.6.1.2)
N
M
Mxyz(,,)
0
Figure 4.34. Plane passing through a point M
0
and perpendicularly to a vector N.
Example 1. Let us write out the equation of the plane that passes through the point M
0
(1, 2, 1)andis
perpendicular to the vector N =(3, 2, 3).
According to (4.6.1.1), the desired equation is 3(x – 1)+2(y – 2)+3(z – 1)=0 or 3x + 2y + 3z – 10 = 0.
4.6.1-2. General equation of plane.
The general (complete) equation of a plane has the form
Ax + By + Cz + D = 0,orr ⋅ N + D = 0.(4.6.1.3)
It follows from (4.6.1.1) that D =–Ax
0
– By
0
– Cz
0
. If one of the coefficients in the
equation of a plane is zero, then the equation is said to be incomplete:
4.6. LINE AND PLANE IN SPACE 125
Figure 4.35. Basis in plane.
1. For D = 0, the equation has the form Ax + By + Cz = 0 and defines a plane passing
through the origin.

2. For A = 0 (respectively, B = 0 or C = 0), the equation has the form By + Cz + D = 0
and defines a plane parallel to the axis OX (respectively, OY or OZ).
3. For A=D=0 (respectively, B =D=0 or B =D =0), theequation has the form By+Cz=0
and defines a plane passing through the axis OX (respectively, OY or OZ).
4. For A = B =0 (respectively, A =C = 0 or B =C = 0), the equation has the form Cz+D =
0
and defines a plane parallel to the plane OXY (respectively, OXZ or OY Z).
4.6.1-3. Parametric equation of plane.
Each vector
−−−→
M
0
M = r – r
0
lying in a plane (where r and r
0
are the position vectors of the
points M and M
0
, respectively) can be represented as (see Fig. 4.35)
−−−→
M
0
M = tR
1
+ sR
2
,(4.6.1.4)
where R
1

=(l
1
, m
1
, n
1
)andR
2
=(l
2
, m
2
, n
2
) are two arbitrary noncollinear vectors lying in
the plane. Obviously, these two vectors form a basis in this plane. The parametric equation
of a plane passing through the point M
0
(x
0
, y
0
, z
0
)hastheform
r = r
0
+ tR
1
+ sR

2
,or
x = x
0
+ tl
1
+ sl
2
,
y = y
0
+ tm
1
+ sm
2
,
z = z
0
+ tn
1
+ sn
2
.
(4.6.1.5)
4.6.1-4. Intercept equation of plane.
AplaneAx + By+Cz+D = 0 that is not parallel to the axis Ox (i.e., A ≠ 0) meets this axis
at a (signed) distance a =–D/A from the origin (see Fig. 4.36). The number a is called the
x-intercept of the plane. Similarly, one defines the y-intercepts b =–D/B (for B ≠ 0)and
the z-intercept c =–D/C (for C ≠ 0). Then such a plane can be defined by the equation
x

a
+
y
b
+
z
c
= 1,(4.6.1.6)
which is called the intercept equation of the plane.
126 ANALYTIC GEOMETRY
O
b
a
X
Y
Z
c
Figure 4.36. A plane with intercept equation.
N
N
γ
β
α
0
X
Y
Z
Figure 4.37. A plane with normalized equation.
Remark 1. Equation (4.6.1.6) can be obtained as the equation of the plane passing through three given
points.

Remark 2. A plane parallel to the axis OX but nonparallel to the other two axes is definedbythe
equation y/b + z/c = 1,whereb and c are the y-andz-intercepts of the plane. A plane simultaneously parallel
to the axes OY and OZ can be represented in the form z/c = 1.
Example 2. Consider the plane given by the general equation 2x + 3y – z + 6 = 0. Let us rewrite it in
intercept form.
The x-, y-, and z-intercepts of this plane are
a =–
D
A
=–
6
2
=–3, b =–
D
B
=–
6
3
=–2,andc =–
D
C
=–
6
–1
= 6.
Thus the intercept equation of the plane reads
x
–3
+
y

–2
+
z
6
= 0.
4.6.1-5. Normalized equation of plane.
The normalized equation of a plane has the form
r ⋅ N
0
– p = 0,orx cos α + y cos β + z cos γ – p = 0,(4.6.1.7)
where N
0
=(cosα,cosβ,cosγ) is a unit vector and p is the distance from the plane to
the origin; here cos α,cosβ,andcosγ are the direction cosines of the normal to the
plane (see Fig. 4.37). The numbers cos α,cosβ,cosγ,andp can be expressed via the
coefficients A, B, C as follows:
cos α =
A
√
A
2
+ B
2
+ C
2
,cosβ =
B
√
A
2

+ B
2
+ C
2
,
cos γ =
C
√
A
2
+ B
2
+ C
2
, p =
D
√
A
2
+ B
2
+ C
2
,
(4.6.1.8)
where the upper sign is taken if D < 0 and the lower sign is taken if D > 0.ForD = 0,
either sign can be taken.
The normalized equation (4.6.1.7) can be obtained from a general equation (4.6.1.3) by
multiplication by the normalizing factor
μ =

1
√
A
2
+ B
2
+ C
2
,(4.6.1.9)
where the sign of μ must be opposite to that of D.
4.6. LINE AND PLANE IN SPACE 127
Example 3. Let us reduce the equation of the plane –2x + 2y – z – 6 = 0 to normalized form.
Since D =–6 < 0, we see that the normalizing factor is
μ =
1

(–2)
2
+ 2
2
+(–1)
2
=
1
3
.
We multiply the equation by this factor and obtain
–
2
3

x +
2
3
y –
1
3
z – 2 = 0.
Hence for this plane we have
cos α =–
2
3
,cosβ =
2
3
,cosγ =–
1
3
, p = 2.
Remark. The numbers cos α,cosβ,cosγ,andp are also called the polar parameters of a plane.
4.6.1-6. Equation of plane passing through point and parallel to another plane.
The plane that passes through a point M
1
(x
1
, y
1
, z
1
) and is parallel to a plane Ax + By +
Cz + D = 0 is given by the equation

A(x – x
1
)+B(y – y
1
)+C(z – z
1
)+D = 0.(4.6.1.10)
Example 4. Let us derive the equation of the plane that passes through the point M
1
(1, 2,–1)andis
parallel to the plane x + 2y + z + 2 = 0.
According to (4.6.1.1), the desired equation is (x – 1)+2(y – 2)+(z + 1)+2 = 0 or
x + 2y + z – 2 = 0.
4.6.1-7. Equation of plane passing through three points.
The plane passing through three points M
1
(x
1
, y
1
, z
1
), M
2
(x
2
, y
2
, z
2

), and M
3
(x
3
, y
3
, z
3
)
(see Fig. 4.38) is described by the equation





x – x
1
y – y
1
z – z
1
x
2
– x
1
y
2
– y
1
z

2
– z
1
x
3
– x
1
y
3
– y
1
z
3
– z
1





= 0,or

(r – r
1
)(r
2
– r
1
)(r
3

– r
1
)

= 0,(4.6.1.11)
where r, r
1
, r
2
,andr
3
are the position vectors of the points M(x, y, z), M
1
(x
1
, y
1
, z
1
),
M
2
(x
2
, y
2
, z
2
), and M
3

(x
3
, y
3
, z
3
), respectively.
M
M
M
Mxyz(,,)
1
2
3
Figure 4.38. Plane passing through three points.
128 ANALYTIC GEOMETRY
Remark 1. Equation (4.6.1.11) means that the vectors
−−−→
M
1
M,
−−−→
M
1
M
2
,and
−−−→
M
1

M
3
are coplanar.
Remark 2. Equation (4.6.1.11) of the plane passing through three given points can be represented via a
fourth-order determinant as follows:







xyz1
x
1
y
1
z
1
1
x
2
y
2
z
2
1
x
3
y

3
z
3
1







= 0.(4.6.1.11a)
Remark 3. If the three points M
1
(x
1
, y
1
, z
1
), M
2
(x
2
, y
2
, z
2
), and M
3

(x
3
, y
3
, z
3
) are collinear, then equa-
tions (4.6.1.11) and (4.6.1.11a) become identities.
Example 5. Let us construct an equation of the plane passing through the three points M
1
(1, 1, 1),
M
2
(2, 2, 1), and M
3
(1, 2, 2).
Obviously, the points M
1
, M
2
,andM
3
are not collinear, since the vectors
−−−→
M
1
M
2
=(1, 1, 0)and
−−−→

M
1
M
3
=
(0, 1, 1) are not collinear. According to (4.6.1.11), the desired equation is





x – 1 y – 1 z – 1
110
011





= 0,
whence
x – y + z – 1 = 0.
4.6.1-8. Equation of plane passing through two points and parallel to line.
The equation of the plane passing through two points M
1
(x
1
, y
1
, z

1
)andM
2
(x
2
, y
2
, z
2
)and
parallel to a straight line with direction vector R =(l, m, n) (see Fig. 4.39) is





x – x
1
y – y
1
z – z
1
x
2
– x
1
y
2
– y
1

z
2
– z
1
lmn





= 0,or

(r – r
1
)(r
2
– r
1
)R

= 0,(4.6.1.12)
where r, r
1
,andr
2
are the position vectors of the points M (x, y, z), M
1
(x
1
, y

1
, z
1
), and
M
2
(x
2
, y
2
, z
2
), respectively.
R
R
M
M
Mxyz(,,)
2
1
Figure 4.39. Plane passing through two points and parallel to line.
Remark. If the vectors
−−−→
M
1
M
2
and R are collinear, then equations (4.6.1.12) become identities.
Example 6. Let us construct an equation of the plane passing through the points M
1

(0, 1, 0)andM
2
(1, 1, 1)
and parallel to the straight line with direction vector R =(0, 1, 1).
According to (4.6.1.12), the desired equation is





x – 0 y – 1 z – 0
1 – 01– 11– 0
011





= 0,
whence
–x – y + z + 1 = 0.