Chapter 12 MATB 113

ADVANCED CALCULUS & ANALYTICAL GEOMETRY

(MATB 113)

CHAPTER 12

VECTORS & THE GEOMETRY OF SPACES

.:SYLLABUS CONTENTS:.

12.1 Three-Dimensional Coordinate Systems

12.2 Vectors

12.3 The Dot Product

12.4 The Cross Product

12.5 Lines and Planes in Space

12.6 Cylinders and Quadric Surfaces

Advanced Calculus & Analytical Geometry ~ MATB 113

Vectors & The Geometry of Spaces 2

12.1 Three Dimensional Coordinate Systems

Learning Objectives:

At the end of this topic students should;

be acquainted with three-dimensional coordinate systems.

be able to interpret equations and inequalities geometrically.

be able to find distance and spheres in space.

be able to find center and radius of a sphere.

Rectangular Coordinate Systems

The coordinate axes, taken in pairs, determine three

coordinate planes: the xy-plane, the xz-plane and the

yz-plane.

To each point P in three dimensional space, we can

assign a triple of real numbers by passing three



plane through P and we denote the point P as

P(a, b, c).



Interpreting Equations and Inequalities

Geometrically

In the following examples, we match coordinate

equations and inequalities with the set of points they

define in space.

(a) 0 1x

(b) 2 , 1z y x

(c) 0, 0, 0z x y

(d) 2 2 1, 3x y z

Distance and Spheres in Space

The formula for the distance between two points in the

xy-plane extends to points in space.

The Distance Between ),,( 1111 zyxP and ),,( 2222 zyxP is

2

12

2

12

2

1221 )()()( zzyyxxPP



We can use the distance formula to write equations for

spheres in space.

The Standard Equation for the Sphere of Radius a and

Center ),,( 000 zyx

2 2 2 2

0 0 0( ) ( ) ( )x x y y z z a

___________________________________________

Example 12.1.1 .:

a) Find the distance between (3, 4, 5) and (5, 3, 1).

b) Show that the graph of the equation

0364222 yxzyx is a sphere, and find

its center and radius.



12.2 Vectors



be acquainted with the definition of vector and the associated notation

and terminology.

be able to represent vectors geometrically as directed line segments or

algebraically as ordered pairs or triples of numbers.

be able to determine the length of a vector.

be able to apply the vector addition and scalar multiplication.

understand the definitions of unit vectors and be able to apply the

definition in terms of finding a vectors direction.

Component Form

A quantity such as force, displacement or

velocity is called a vector and is represented by

a directed line segment.

The arrow points in the direction of the action

and its length gives the magnitude of the action

in terms of a suitably chosen unit.



For example, a force vector points in the

direction in which the force acts while its length

is a measure of the forces strength.

Definition:(Vector, Initial and Terminal Point, Length)

A vector in the plane is a directed line segment.

The directed line segment

AB has initial point A and

terminal point B. Its length is denoted by

AB . Two

vectors are equal if they have same length and

direction.

.B terminal point

AB A. Initial point



Vectors are usually written in lowercase and

boldface letters( u, v and w).

In handwritten form, it is customary to draw

small arrows above or below the letters

(

u and ~v ).

Definition:(Component Form)

If v is a two-dimensional vector in the plane equal to the

vector with initial point at the origin and terminal point

(v1, v2), then the component form of v is

21,vvv

If v is a three-dimensional vector in the plane equal to

the vector with initial point at the origin and terminal

point (v1, v2, v3), then the component form of v is

321 ,, vvvv



Norm of a Vector

The distance between the initial and terminal points

of a vector v is called the length, the norm, or the

magnitude of v and is denoted by v or v .

The distance does not change if the vector is

translated, so purposes of calculating the norm we

can assume that the vector is positioned with its

initial point at the origin.



The magnitude or length of the vector 21,vvv in

two-dimensional space is,

1 2

2 2v v v

while, for the vector 321 ,, vvvv in three-

dimensional space, the magnitude is,

1 2 3

2 2 22 2 2

2 1 2 1 2 1v v v x x y y z z v

Example 12.2.1:

Given a vector with initial point P(-3, 4, 1) and

terminal point Q(-5, 2, 2).Find,

a) the component form of the vector,

b) the length of the vector.



Vector Algebra Operations

Two principal operations involving vectors are

vector addition and scalar multiplication.

A scalar is simply a real number, and is called

such when we want to draw attention to its

difference from vectors. Scalar can be positive,

negative or zero.

Definition:(Vector Addition and Scalar Multiplication)

Let 321 ,, uuuu and 321 ,, vvvv be vectors with

k a scalar.

Addition : 1 1 2 2 3 3, ,u v u v u v u v

Scalar multiplication: 1 2 3, ,k ku ku ku u

The definition of a vector addition is illustrated

geometrically for planar vectors like in the



figure below, where the initial of one vector is

placed at the terminal point of the other.

Another interpretation is shown like in the

figure below, where the sum, called the

resultant vector, is the diagonal of the

parallelogram.



Example 12.2.2:

Let 3, 2 u and 2,5 v . Find the component

form and the magnitude of the vector

a) 2 v b) 2 3u v



Example 12.2.3:

Copy vectors u, v and w head to tail as needed to

sketch the indicated vector.

u w

v

a) vu b) wvu

c) vu d) vu

Properties of vector Operations

Let u,v,w be vectors and a, b are scalars

1. u + v = v + u 2. (u + v) + w = u + (v + w)

3. u + 0 = u 4. u + (-u) = 0

5. 0u = 0 6. 1u = u

7. a (b u) = (ab) u 8. a (u + v) = a u + a v

9. (a + b) u = a u + b u



Unit Vectors

A unit vector is a vector of magnitude 1.

The vectors i, j and k are unit vectors.

The standard unit vectors are

1,0,0 , 0,1,0 , i j and 0,0,1k

Whenever 0v , then the unit vector u that has the

same direction as v is

v

uv

Note : If 321 ,, uuuu , then 1 2 3u u u u i j k

Example 12.2.4:

Find the unit vector u that has same direction as the

vector 3i-4j.



Vectors Determined by Length and a Vector in the

Same Direction

It is a common problem in many applications that

a direction in 2-space or 3-space is determined by

some known unit vector u, and it is of interest to

find the components of a vector v that has the

same direction as u and some specified length v .

This can be done by expressing v as

uvv

v is equal to its length times a unit vector in the same direction

and reading off the components of uv .

Example 12.2.5:

a) Find the vector with the same direction and one-

half the magnitude of 4i-6j.



b) Find the unit vector with the direction opposite

to -3i+4j.

c) Find the unit vector that makes an angle

3 / 4 with the positive x- axis.

Write your answer in the component form.



Midpoint of a Line Segment

Vectors are often useful in geometry. For example,

the coordinates of the midpoint of a line segments

are found by averaging.

The midpoint M of the line segment joining points

),,( 1111 zyxP and ),,( 2222 zyxP is the point,

2,

2,

2

212121 zzyyxx

Example 12.2.6:

Find the midpoint of the line segment from (2,-3,6) to

(3,4,-2).



12.3 The Dot Product



understand and be able to use the definitions of the dot product to

measure the length of a vector and the angle between two vectors.

understand and be able to apply properties of the dot product.

understand that two vectors are orthogonal /perpendicular if their dot

product is zero.

understand be able to determine the projection of a vector a onto a

vector b.

be able to use vectors and the dot product in many applications.

The Dot Product

The dot product is also known as a scalar product or

inner product because it is a product of vectors that

gives a scalar (that is, real number) as a result.

Definition:(The Dot Product)

The dot product vu of a vectors 321 ,, uuuu and

321 ,, vvvv is

332211 vuvuvu vu



Example 12.3.1

Find the dot product of kjiu 23 and

kjiv 24 .

Example 12.3.2

If 3,1,4v and 5,2,1w , find the dot

product, wv .

Before we can apply the dot product to geometric

and physical problems, we need to know how it

behaves algebraically.

A number of important general properties of the dot

product are listed in the following table.



Properties of the dot product.

If u, v and w are vectors in 3 and c is a scalar, then

Magnitude of a vector 2

vvv

Zero Product 0v0

Commutativity vwwv

Multiple of a dot product )()()( wvwvwv ccc

Distributivity wuvuwvu )(

Angle Between Vectors

When two nonzero vectors u and v are placed so

their initial points coincide, they form and angle

of a measure 0 as shown in figure.



Definition:(Angle Between Vectors)

The angle between two nonzero vectors

321 ,, uuuu and 321 ,, vvvv is given by

1cos

u v

u v

If is the angle two nonzero vectors u and v , then

(a) is an acute angle if and only if 0vu .

(b) is an obtuse angle if and only if 0vu

(c) 090 is a right angle if an only if 0vu

Example 12.3.3:

a) Find the angle between kjiu 23 and kiv .



b) Given the triangle ABC with vertices A(4, 1, 2),

B(3, 4, 5) and C(5, 3, 1). Find the acute angle CAB .

c) The angle between vector u and v is

21

4cos 1 .

Find the scalar t given that,

kjiu 236 and kjiv 42 t

Perpendicular/Orthogonal Vectors

Definition:(Perpendicular Vectors)

Two vectors u and v are said to be orthogonal

(perpendicular) to each other if and only if

i) at least one of them is a null vector, or

ii) the angle between them is a right angle

so that 0vu .

In other words, the vector u is orthogonal to v

iff 0vu



Example 12.3.4:

Determine the value of so that kjiu 2 and

kjiv 224 are orthogonal.

Vector Projection

Wooden block

1w 2w

Smooth slope F

Fig.12.1

F = the weight of the wooden block

= 21 ww



Projection and Vector Components

Definition:(Projection and Vector Components)

Let u and v be a non-null vectors. If 21 wwu with 1w

parallel to v , and 2w orthogonal to v , then

i) 1w is the projection of u onto v or vector component u

along v , and is labeled as uw vproj1 .

ii) 12 wuw is the vector component of u

perpendicular to v .

Acute Angle



Obtuse Angle

Fig.12.2: The Vector Projection of u onto v

Projection Using Dot Product

Let u and v be a non-null vectors and 1w , 2w be the

component of u parallel and perpendicular to v , then

1. v

v

vuuw v

21proj

(Vector component of u in the direction of v )

2. uuwuw vproj 12

(Vector component of u orthogonal to v )



3. v

vu

v

vuuuv

cosproj

(Scalar component of u in the direction of v )

Example 12.3.5:

Given the points A(2, 2, 3), B(-1, 5, 4) and C(3, -5, 1).

Find the projection vector

AB in the direction

AC .

Example 12.3.6:

Given 2 2 u i j k and 2 10 11 v i j k . Find the scalar

component of u in the direction of v .



Work done by a Force as a Scalar Product

W = (force x displacement)

Fd

If F makes an angle with the direction of the motion

of the particle, then the definition of work done is



Example 12.3.7:

A person pulls a wagon along level ground by exerting a

force of 20 pounds on a handle that makes an angle of

030 with the horizontal. Find the work done in pulling

the wagon100 feet.

How to write u as a vector parallel to v plus a vector

orthogonal to v

u = projvu + (u projvu)

=

v

v

vuuv

v

vu22

..



12.4 The Cross Product



understand and be able to use the definitions of the cross product to find

vectors that orthogonal to each other.

understand that two vectors are parallel if and only if their cross product

is zero.

understand and be able to apply properties of the cross product.

be able to use vectors and the cross product in many applications.

The Cross Product

We start with two nonzero vectors u and v in space.

If u and v are not parallel, they determine a plane.

We select a unit vector n perpendicular to the plane

by the right-hand rule.



This means that we choose n to be the unit (normal)

vector that points the way your right thumb points

when your finger curl through the angle from u

to v.

Unlike the dot product, the cross product is a

vector.

For this reason its also called the vector product

of u and v.

The vector vu is orthogonal to both of u and v

because it is a scalar multiple of n.

Definition: (Cross Product)

sin u v u v n

Nonzero vectors u and v are parallel if and only if

0 vu .

The vector vu is orthogonal to both u and v



Determinant Formula for vu

If kjiu 321 uuu and kjiv 321 vvv , then

1 2 3

1 2 3

2 3 1 3 1 2

2 3 1 3 1 2

2 3 3 2 1 3 3 1 1 2 2 1( ) ( ) ( )

u u u

v v v

u u u u u u

v v v v v v

u v u v u v u v u v u v

i j k

u v

i j k

i j k

Example 12.4.1:

a) If kjia 2 and kjib 2 , calculate ba .

b) Find the unit vector orthogonal to the vectors

kjia 4 and kjib 22 .



Properties of the Cross Product

If u, v and w are vectors in 3 and r, s are scalar, then

( ) ( ) ( )r s rs u v u v

)()()( wuvuwvu

( ) v u u v

( ) ( ) ( ) v w u v u w u

0 0 u

( ) ( ) ( ) u v w u w v u v w



vu is the Area of a Parallelogram

Because n is a unit vector, the magnitude of vu is

sin

sin

u v u v n

u v

This is the area of the parallelogram determined by u

and v.

u be the base of the parallelogram and sinv is the

height.

Remarks: If a

OA and b

OB , then

i) the area of the parallelogram OACB is ba

OBOA.

ii) the area of the triangle OAB is ba

2

1

2

1OBOA

.



Example 12.4.2:

a) Find the area of parallelogram bounded by the

vectors kjia 4 and kjib 32 .

b) Find the area of the triangle with vertices

P(4, -3, 1), Q(6, -4, 7)and R(1, 2, 2).



Triple Scalar Product or Box Product

The product wvu )( is called the triple scalar

product of u, v and w.

The results of the process is a scalar.

Geometrically, this product is the volume of the

parallelepiped determined by three vectors given.

By treating the planes of v and w and of w and u as

the base planes of the parallelepiped determined by

u, v and w, we see that

vuwuwvwvu )()()(



Since the dot product is commutative, we also have

)()( wvuwvu .

Calculating the Triple Scalar Product as a

Determinant

The triple scalar product can be evaluated as a

determinant:

312212133112332

321

321

321

)()()(

)(

wvuvuwvuvuwvuvu

vvv

uuu

www

wvu

Example 12.4.3:

a) Find the volume of the box(parallelepiped)

determined by kjiu 2 , kiv 2 and

kjw 47 .



b) Find the volume of the parallelepiped with adjacent

edges

OP ,

OQ and

OR where P(1, 3, -2),

Q(2, 4, 5)and R(-3, -2, 2).

c) Use the triple scalar products to show that the points

P(2, 0, 1), Q(3, 2, 0), R(1, -1, 2). and S(5, 4, -2) are

coplanar.

(Coplanar = points that lie within the same plane)



Torque

The torque vector points in the direction of the axis of

the bolt according to the right-hand rule (clockwise from

tip of the vector)

Magnitude of torque vector = sin r F r F

Let n be the unit vector along the bolt axisin direction of

the torque, then

Torque vector = sinr F n

0 vu when u and v are parallel



12.5 Lines and Planes in Space



understand and be able to describe lines and planes by using the vector

concepts of parallel and orthogonal, respectively.

be able to find an equation of straight line and plane in space.

be able to calculate angle between two intersecting lines.

be able to find the shortest distance from a point to a line.

be able to calculate the angle between two planes.

be able to find the shortest distance of a point from a plane.

be able to find the line of intersection of two planes.

Lines in Space

In the plane, a line is determined by a point and a

number giving the slope of the line.

In space, a line is determined by a point and vector

giving the direction of the line.



Suppose L is a straight line that passes through the

point ),,( 0000 zyxP and is parallel to the vector

kjiv 321 vvv .

Then, another point P( x, y, z) lies on the L if and

only if the vectors v and

PP0 are parallel, that is

vtPP

0

for a real number of t.

If

00 OPr and

OPr are the position vectors of

the points P0 and P respectively, then

00 rr

PP



Hence,

vr

vrr

t

t

0

0

, (represents the line L)

We can write the above expression in component

form as

321000 ,,,,,, vvvtzyxzyx

Equating the components and solving for x, y and z

gives,

302010 tvzztvyytvxx

where t is a real number. These are parametric

equations for the line L, with parameter t.



Parametric Equations for a Line

The standard parametrization of the line through

),,( 0000 zyxP to 1 2 3v v v v i j k is

0 1 0 2 0 3x x tv y y tv z z tv

Example 12.5.1:

a) Write down the parametric equations for the

straight line passing through point P(2, 3, 5) and

parallel to 2 v i j k .

b) Find the parametric equations for the line passing

through the points P(0, 8, 4) and Q(2, 4, 5).



Example 12.5.2:

a) Find the parametric equations for the line L

through P(5, -2, 4) that is parallel to

3

2,2,

2

1a .

b) Where does L intersect the xy-plane?



The Distance from a Point to a Line in Space

Let P be a point on a line L and let v be a vector

parallel to L.

The shortest distance from a point S to the line L is

given by

sind PS

where is the angle between v and vector

PS .



Since sin

PSPS vv, therefore we have the

shortest distance of S from line L as

sind PS

PS

v

v

Example 12.5.3:

a) Find the shortest distance from the point

S(1, 0, -1) to the line,

tztytxL 21132:

b) Find the shortest distance from the point

S(5, 2, -1) to the line : (2 3 )L t r i i j k



Plane in Space

Suppose that M is a plane, where on it lies a point

),,( 0000 zyxP with its position vectors

0000 ,, zyxr .

Let P( x, y, z) be any point on M with its position

vectors zyx ,,r .

So, the vector

00000 ,, zzyyxxPP rr

lies on M.



If kjin cba is a non-null vector orthogonal to

M, then n is orthogonal to

PP0 , that is,

0)()()(

0,,,,

0

000

000

0

zzcyybxxa

cbazzyyxx

PP n

Equations for a Plane

The plane through ),,( 0000 zyxP normal to

kjin cba has,

Vector Equation : 00

PPn

Component Equation :

0 0 0( ) ( ) ( ) 0A x x B y y C z z

Component Equation Simplified: Ax By Cz D ,

where 0 0 0D Ax By Cz



Example 12.5.4:

a) Find the equation of a plane that contains the point

P(5, -2, 4) and the normal vector 1,2,3n .

b) Find the equation of the plane that contains of the

points P(-1, 2, 1), Q(0, -3, 2) and R(1, 1, -4).

c) Find the equation of the plane that is

perpendicular to plane 0 zyx and

042 zyx and passing through the point

(4, 0, -2).



Lines of Intersection

When a Plane M1 intersects another Plane M2, we

obtain a line L.

The coordinates of every points on the line L will

satisfy the equations of both these planes.

To obtain the equation of the line of intersection of

two planes, we need

a) a vector parallel to the line L which is given

by 21 nn .

b) A point ),,( 000 zyx on the line L that can be

chosen by solving the equations of the planes.



If cba ,,21 nn , then the equation of the line L

in a parametric form is given by

0 1 0 2 0 3x x tv y y tv z z tv

Example 12.5.5:

Find the equation of a line that passes through (-1,2,3)

and is parallel to the line of intersection between the

planes 323 zyx and 52 zyx .



The Distance from a Point to a Plane

If P is a point on a plane with normal n, then the

distance from any point S to the plane is the length of

the vector projection of PS

onto n. That is, the

distance from S to the plane is

d PS

n

n

where )A B C n i j k is normal to the plane .



Angle Between Planes

The angle between two intersecting planes is

defined to be the angle between their normal vectors

.

The angle between plane M1 intersects another plane

M2 is equal to the angle between normal vector n1

and n2. If is the acute angle between the two

planes, then

1 1 2

1 2

cos

n n

n n



In here we conclude that,

i) the angle between two intersecting planes is the

angle between the normal vectors to the planes.

ii) Two planes are parallel if and only if 21 nn ,

for a certain .

iii) Two planes are orthogonal if and only if

021 nn .

Example 12.5.6:

a) Find the angle between the planes 2 zyx

and 342 zyx .

b) Show that the planes 532 zyx and

2396 zyx are parallel.



12.6 Cylinders and Quadric Surfaces

Recall :



Cylinders

A cylinder is a surface that is generated by moving a

straight line along a given planar curve while

holding the line parallel to a given fixed line.

The curve is called a generating curve for the

cylinder.



In solid geometry, where cylinder means circular

cylinder, the generating curve are circles, but now

we allow generating curves of any kind.



Quadric Surfaces

A quadric surface is the graph in space of a second-

degree equation in x, y and z.

The most general form is,

0222 KJzHyGxFxzEyzDxyCzByAx

The basic quadric surfaces are ellipsoid, paraboloid,

cones and hyperboloids.



Ellipsoid

The equation

12

2

2

2

2

2

c

z

b

y

a

x

is an ellipsoid and its graph is as shown below



Some traces of the ellipsoid 12

2

2

2

2

2

c

z

b

y

a

x is shown

in table,

Traces Equation of Trace Graph

xy-trace 12

2

2

2

b

y

a

x Ellipse

xz-trace 12

2

2

2

c

z

a

x Ellipse

yz-trace 12

2

2

2

c

z

b

y Ellipse

If a = b = c, then the equation reduces to

2222 azyx and it is a sphere of radius a.



Example 12.6.1:

Sketch the graphs of each question in three dimensions.

a) 2xy b) 9

22 zx c) 922 xy

Example 12.6.2:

Sketch the surfaces in three dimensional.

a) 99222 zyx



Hyperboloid of One Sheet

The equation

12

2

2

2

2

2

c

z

b

y

a

x

is a hyperboloid of one sheet and its graph is as shown

below,



Some traces of the hyperboloid of one sheet

12

2

2

2

2

2

c

z

b

y

a

x is shown in table below,


xy-trace 12

2

2

2

b

y

a

x Ellipse

xz-trace 12

2

2

2

c

z

a

x Hyperbola

yz-trace 12

2

2

2

c

z

b

y Hyperbola

Example 12.6.3:

Sketch the surfaces in three dimensional.

a) 1222 zyx



Hyperboloid of Two Sheets

The equation

12

2

2

2

2

2

c

z

b

y

a

x

is a hyperboloid of two sheets and its graph is as

shown below :



Some traces of the hyperboloid of two sheets

12

2

2

2

2

2

c

z

b

y

a

x is shown in table below,


xy-trace 12

2

2

2

b

y

a

x No graph

xz-trace 12

2

2

2

c

z

a

x Hyperbola

yz-trace 12

2

2

2

c

z

b

y Hyperbola

Example 12.6.4:

Sketch the graph of the equations in three dimensional.

a) 99222 zxy



Cone

The equation

02

2

2

2

2

2

c

z

b

y

a

x

is a double cone and its graph is as shown in figure,



Some traces of the cone 02

2

2

2

2

2

c

z

b

y

a

x is shown in

table below :


xy-trace 02

2

2

2

b

y

a

x The origin

xz-trace 02

2

2

2

c

z

a

x The lines

xa

cz

yz-trace 02

2

2

2

c

z

b

y The lines

yb

cz

The axis of the cone is the z-axis. The trace in a

plane 0zz parallel to the xy-plane had the equation

2

2

0

2

2

2

2

c

z

b

y

a

x

Example 12.6.5:


a) 2 2 24x y z



Paraboloid

The equation

0,2

2

2

2

cczb

y

a

x

is a paraboloid and its graph is as shown in the

following figure,

Some traces of the paraboloid czb

y

a

x

2

2

2

2

is

shown in following table :



Traces Equation of

Trace Graph

xy-trace 12

2

2

2

b

y

a

x Ellipse

xz-trace 2

2

cz x

a Parabola

yz-trace 2

2

cz y

b Parabola

The traces in planes parallel to the xy-planes

02

2

2

2

czb

y

a

x

are ellipses.

The axis of the paraboloid is z-axis and its vertex is

the origin.

If c < 0, then the paraboloid opens downward.

If a = b, then the paraboloid is called circular

paraboloid, and traces in planes parallel to the xy-

planes are circles.



Example 12.6.6:


a) zyx 22 4

A Saddle Point (Hyperbolic Paraboloid)

The equation

0,2

2

2

2

cczb

y

a

x

is a hyperbolic paraboloid and its graph is as shown in

the following figure :



A hyperbolic paraboloid is the most difficult

quadric surface to visualize. The trace in the

xy-plane with the equation

02

2

2

2

b

y

a

x or xa

by

is a pair of intersecting line through the origin.

The xz-trace is the parabola

2

2

xcz

a

which assumes maximum value at the origin,

whereas the yz-trace is the parabola

2

2

ycz

b

which assumes minimum value at the origin.

Example 12.6.7:


a) zxy 22

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Chapter 12 MATB 113