mkmm 1213 advanced engineering...

Faculty of Mechanical EngineeringEngineering Computing Panel

MKMM 1213 Advanced Engineering Mathematics

Vectors and the Geometry of Spaces

Abu Hasan Abdullah

April 2015

noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 1 / 57

Outline I

1 Three Dimensional Coordinate Systems

The Cartesian Coordinate System

Distance and Spheres in Space

2 Vectors

Component Form

Vector Algebra Operations

Unit Vectors

Position Vectors

Midpoint of a Line Segment

3 The Dot Product

Angle Between Vectors

Orthogonal Vectors

Dot Product Properties and Vector Projections

4 The Cross Product

The Cross Product of Two Vectors in Space

Properties of the Cross Product

Area of a Parallelogram

Determinant Formula for u × v

Torque

Triple Scalar or Box Productnoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 2 / 57

Outline II

5 Lines and Planes in Space

Lines and Line Segments in Space

Distance from a Point to a Line in Space

Equation for a Plane in Space

Lines of Intersection

Distance from a Point to a Plane

Angles Between Planes

6 Cylinders and Quadric Surfaces

Cylinders

Quadric Surfaces

7 Bibliography


Three Dimensional Coordinate SystemsThe Cartesian Coordinate System

To locate a point in space, we use three

mutually perpendicular coordinate axes,

arranged as in Figure 1.

The axes shown there make a right-handed

coordinate frame.

Looking down on the xy-plane from the

positive direction of the z-axis, positive

angles in the plane are measured

counterclockwise from the positive x-axis

and around the positive z-axis.

The Cartesian coordinates (x, y, z) of a

point P in space are the values at which the

planes through P perpendicular to the axes

cut the axes.

Cartesian coordinates for space are also

called rectangular coordinates because the

axes that define them meet at right angles.

Figure 1: The Cartesian coordinatesystem is right-handed.


Three Dimensional Coordinate SystemsThe Cartesian Coordinate System

The planes in Figure 2 are determined bythe coordinates axes:

the xy-plane (blue), whose standardequation is z = 0;the yz-plane (green), whose standardequation is x = 0; andthe xz-plane (brown), whose standardequation is y = 0.

and meet at the origin (0, 0, 0), which is

identified by 0 or sometimes the letter O.

The three coordinate planes x = 0, y = 0,

and z = 0 divide space into eight cells

called octants. The octant in which the

point coordinates are all positive is called

the first octant; there is no convention for

numbering the other seven octants.

Figure 2: The planes x = 0, y = 0, andz = 0 divide space into eight octants.


Three Dimensional Coordinate SystemsDistance and Spheres in Space

The formula for the distance between two points in 2-D plane extends to points in

3-D space, Figure 3. Hence, the distance between P1 (x1, y1, z1) and P2(x2, y2, z2) is

|P1P2| =p

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2 (1)

Figure 3: Distance between P1 and P2 by applying the Pythagorean theorem.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 6 / 57

Three Dimensional Coordinate SystemsDistance and Spheres in Space

Eq. 1 can be used to write equation for a sphere in space, Figure 4. A point P(x, y, z)lies on the sphere of radius a centered at P0(x0, y0, z0) precisely when |P0P| = a or

(x − x0)2 + (y − y0)

2 + (z − z0)2 = a

2(2)

Figure 4: Sphere of radius a centred at the point x0, y0, z0).noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 7 / 57

VectorsComponent Form

DEFINITION:

The vector represented by the di-

rected line segment⇀

AB, Figure 5,

has initial point A and terminal

point B and its length is denoted

by |⇀

AB|.

Figure 5: The directed line segment⇀

AB iscalled a vector.

Two vectors are equal if they have

the same length and direction, as

shown in Figure 6.

Figure 6: Equal vectors⇀

AB =⇀

CD =⇀

OP =⇀

EF.

Vectors are used to represent things

that have both magnitude and

direction in the plane or in space.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 8 / 57

VectorsComponent Form

DEFINITION:

Let v =⇀

PQ; see Figure 7.If v is a two-dimensional vector in theplane equal to the vector with initialpoint at the origin and terminal point(v1, v2), then the component form of vis

v = 〈v1, v2〉

If v is a three-dimensional vector in thespace equal to the vector with initialpoint at the origin and terminal point(v1, v2, v3), then the component formof v is

v = 〈v1, v2, v3〉

Figure 7: A vector v =⇀

PQ in standardposition has its initial point at the origin.

Note: The only vector with length 0 is the zero vector

0 = 〈0, 0〉 or 0 = 〈0, 0, 0〉. This vector is also the only

vector with no specific direction.

The magnitude or length of vector v =⇀

PQ is the non-negative number:

|v| =q

v21 + v2

2 + v23 =

p

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 9 / 57

VectorsVector Algebra Operations

Two principal operations involving vectors are vector addition and scalar

multiplication.

A scalar is simply a real number. Scalars can be positive, negative, or zero and are

used to scale a vector by multiplication.

DEFINITION:

Let u = 〈u1, u2, u3〉 and v = 〈v1, v2, v3〉 be vectors with k a scalar

Addition

u + v = 〈u1 + v1, u2 + v2, u3 + v3〉

Scalar multiplication

ku = 〈ku1, ku2, ku3〉



Figure 8: Vector addition (a) geometric interpretation, (b) the parallelogram law.

In Figure 8(a) the initial point

of one vector is placed at the

terminal point of the other.

The parallelogram law of vector addition is

shown in Figure 8(b) where the sum, called

the resultant vector, is the diagonal of the

parallelogram.

In engineering, forces, velocities and accelerations add vectorially, e.g. the force

acting on a particle subject to two gravitational forces is obtained by adding the

two force vectors.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 11 / 57


Figure 9 displays a geometric

interpretation of the product ku of

the scalar k and vector u. If k > 0,

then ku has the same direction as u;

if k < 0, then the direction of ku is

opposite to that of u.

Figure 9: Scalar multiples of u.

Comparing the lengths of u and ku,

we see that

|ku| =p

(ku1)2 + (ku2)2 + (ku3)2

=q

k2(u21 + u2

2 + u23)

=√

k

q

u21 + u2

2 + u23

= |k||u|

The length of ku is the absolute value

of the scalar k times the length of u.

The vector (−1)u = −u has the same

length as u but points in the opposite

direction.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 12 / 57


If u = 〈u1, u2, u3〉 and v = 〈v1, v2, v3〉,then

u − v = 〈u1 − v1, u2 − v2, u3 − v3〉

Note that (u − v) + v = u, so adding

the vector (u − v) to v gives u, see

Figure 10(a).

Figure 10(b) shows the difference

u − v as the sum u + (−v).

Figure 10: (a) The vector u − v, when addedto v, gives u. (b) u − v = u + (−v).noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 13 / 57

VectorsProperties of Vector Operations

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

u + v = v + u (u + v) + w = u + (v + w)

u + 0 = u u + (−u) = 0

0u = 0 1u = u

a(bu) = (ab)u a(u + v) = au + av

(a + b)u = au + bu

When three or more space vectors lie in the same plane, we say they are coplanar

vectors. For example, the vectors u, v, and u + v are always coplanar.


VectorsUnit Vectors

A vector v of length 1 is called a unitvector. The standard unit vectors are

i = 〈1, 0, 0〉 , j = 〈0, 1, 0〉 , k = 〈0, 0, 1〉

Vector v = 〈v1, v2, v3〉 can be writtenas a combination of standard unitvectors:

v = 〈v1, v2, v3〉

= 〈v1, 0, 0〉 + 〈0, v2, 0〉 + 〈0, 0, v3〉

= v1 〈1, 0, 0〉 + v2 〈0, 1, 0〉

+ v3 〈0, 0, 1〉

= v1i + v2j + v3k

where v1, v2 and v3 are the i-, j- and

k-component of v, respectively.

v/|v| is a unit vector in the direction

of v, called the direction of the

nonzero vector v.

In component form, the vector fromP1(x1, y1, z1) to P2(x2, y2, z2), inFigure 11, is

⇀

P1P2 = (x2 − x1)i + (y2 − y1)j

+ (z2 − z1)k

Figure 11: The vector from P1 to P2.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 15 / 57

VectorsPosition Vectors

Example: Find in terms of a, b, the

position vector of the point dividing

the line AB in the ratio m : n, where a

is the position vector of A and b is

the position vector of B with respect

to an origin O.

Figure 12: The position vector of a point.

Solution: If D is the point on AB such that AD/DB = m/n, then

AD =

„

m

m + n

«

AB so⇀

AD =

„

m

m + n

«

⇀

AB

Now

⇀

AB = b − a so⇀

AD =

„

m

m + n

«

b − anoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 16 / 57

VectorsPosition Vectors

Solution (continued):

⇀

OD =⇀

OA +⇀

AD = a +

„

m

m + n

«

(b − a) =na + mb

m + n

If A is the point (x1, y1) and B the point (x2, y2) so that

a = x1i + y1j and b = x2i + y2j

then

⇀

OD =n(x1i + y1j) + m(x2i + y2j)

m + n=

„

nx1 + mx2

m + n

«

i +

„

ny1 + my2

m + n

«

j

i.e. D is the point

„

nx1 + mx2

m + n,

ny1 + my2

m + n

«


VectorsMidpoint of a Line Segment

The midpoint of a line segment are

found by averaging.

The midpoint M of the line segment

joining points P1(x1, y1, z1) and

P2(x2, y2, z2) is the point

“ x1 + x2

2,

y1 + y2

2,

z1 + z2

2

”

Observe in Figure 13 that

⇀

OM =⇀

OP1 + 12(

⇀

P1P2)

=⇀

OP1 + 12(

⇀

OP2 −⇀

OP1)

= 12(

⇀

OP2 +⇀

OP1)

=x1 + x2

2i +

y1 + y2

2j +

z1 + z2

2k

Figure 13: The coordinates of the midpoint arethe averages of the coordinates of P1 and P2.


VectorsMidpoint of a Line Segment: Centroid of a triangle

Example: Consider a triangle ABC

where a, b, c are the position vectors

of A, B, C, respectively. The centroid,

G, of this triangle is the point of

intersection of its medians, i.e. G is

the point which divides CD in the

ratio 2 : 1, D being the midpoint of

AB.

Figure 14: The position vector of the centroidof a triangle.

Hence

⇀

OG =2

⇀

OD +⇀

OC

3and

⇀

OD =a + b

2

Therefore

⇀

OG = 13(a + b + c)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 19 / 57

The Dot ProductAngle Between Vectors

How does one calculate the angle

between two vectors directly from

their components?

Example: If a force F is applied to a

particle moving along a path, we

often need to know the magnitude of

the force in the direction of motion.

If v is parallel to the tangent line to

the path at the point where F is

applied, then we want the magnitude

of F in the direction of v. Figure 15

shows that the scalar quantity we

seek is the length |F| cos θ, where θ is

the angle between the two vectors F

and v.

Figure 15: The magnitude of the force F in thedirection of vector v is the length F cos θ of theprojection of F onto v.



When two nonzero vectors u and v

are placed so their initial points

coincide, they form an angle θ of

measure 0 ≤ θ ≤ π, see Figure 16.

Figure 16: The angle between u and v.

Theorem 1:

Angle Between Two VectorsThe angle θ between two nonzerovectors u = 〈u1, u2, u3〉 andv = 〈v1, v2, v3〉 is given by

θ = cos−1

„

u1v1 + u2v2 + u3v3

|u||v|

«

DEFINITION:

The dot product u ·v (“u dot v”) of

vectors u = 〈u1, u2, u3〉 and

v = 〈v1, v2, v3〉 is the scalar

u · v = u1v1 + u2v2 + u3v3

Hence,

θ = cos−1

„

u · v

|u||v|

«noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 21 / 57


Proof of Theorem 1:Applying the law of cosines to thetriangle in Figure 17,

|w|2 = |u|2 + |v|2 − 2|u||v| cos θ

we find that

2|u||v| cos θ = |u|2 + |v|2 − |w|2

Because w = u − v, component formof w is 〈u1 − v1, u2 − v2, u3 − v3〉. So

|u|2 =

„

q

u21 + u2

2 + u23

«2

= u21 + u2

2 + u33

|v|2 =

„

q

v21 + v2

2 + v23

«2

= v21 + v2

2 + v33

|w|2 = . . .

and

|u|2 + |v|2 − |w|2 = . . .

Therefore

2|u||v| cos θ = . . .

|u||v| cos θ = . . .

cos θ = . . .

Since 0 ≤ θ ≤ π, we have

θ = cos−1 . . .

Figure 17: The parallelogram law of additionof vectors gives w = u − v.

The angle θ between two nonzeronoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 22 / 57

The Dot ProductOrthogonal Vectors

Two nonzero vectors u and v are

perpendicular if the angle between

them is π/2. For such vectors, we

have u · v = 0 because cos π/2 = 0.

The converse is also true. If u and v

are nonzero vectors with

u · v = |u||v| cos θ = 0, then cos θ = 0

and θ = cos−1 0 = π/2.

DEFINITION:

Vectors u and v are orthogonal if u · v = 0.


The Dot ProductDot Product Properties and Vector Projections

The following “dot product” properties hold, if u, v, w are vectors and c is scalar:

u · v = v · u Property 1

(cu) · v = u · (cv) = c(u · w) Property 2

u · (v + w) = u · v + u · w Property 3

(u · u) = |u|2 Property 4

0 · u = 0 Property 5

Proof of Property 1

u · v = u1v1 + u2v2 + u3v3 = v1u1 + v2u2 + v3u3

= v · u

Proof of Property 3

u · (v + w)

= 〈u1, u2, u3〉 · 〈v1 + w1, v2 + w2, v3 + w3〉

= u1(v1 + w1) + u2(v2 + w2) + u3(v3 + w3)

= u1v1 + u1w1 + u2v2 + u2w2 + u2v3 + u3w3

= (u1v1 + u2v2 + u3v3) + (u1w1 + u2w2 + u3w3)

= u · v + u · wnoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 24 / 57


The vector projection of u =⇀

PQ onto

a nonzero vector v =⇀

PS, is vector⇀

PR,

determined by dropping a

perpendicular from Q to line PS, see

Figure 18.

Figure 18: The parallelogram law ofaddition of vectors gives w = u − v.

The notation for the vector projection

of u onto v is

projvu

If u represents a force, then projvu

represents the effective force in the

direction of v, see Figure 19.

Figure 19: The parallelogram law of additionof vectors gives w = u − v.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 25 / 57


If the angle θ between u and v is

acute, projvu has length |u| cos θ and

direction v/|v|, see Figure 20.

If the angle θ between u and v is

obtuse, cos θ < 0 and projvu has

length −|u| cos θ and direction

−v/|v|.

Figure 20: The length of projvu is (a)|u| cos θ if θ is acute, and (b) −|u| cos θ ifθ is obtuse.

In both cases the vector projection ofu onto v is the vector

projvu = (|u| cos θ)v

|v|

=

„

u · v

|v|

«

v

|v|

=

„

u · v

|v|2

«

v

The scalar component of u in the

direction of v is the scalar

|u| cos θ =|u||v| cos θ

|v|=

u · v

|v| = u · v

|v|noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 26 / 57

The Dot ProductWork

A constant force of magnitude F

moving an object through a distance

d is W = Fd. This formula holds only

if the force is directed along the line

of motion.

If a force F moving an object through

a displacement D =⇀

PQ has some

other direction, the work is

performed by the component of F in

the direction of D. If θ is the angle

between F and D, Figure 21, then

Work = (|F| cos θ)|D|= F · D

Figure 21: The work done by a constant forceF during a displacement D.

DEFINITION:

The work done by a constant force

F acting through a displacement

D =⇀

PQ is

W = F · Dnoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 27 / 57

The Dot ProductExamples

Example 1:Find the angle θ in the triangle ABCdetermined by the vertices A = (0, 0),B = (3, 5), and C = (5, 2) (Figure 22).

Figure 22: Example 1.

Example 2:Find the angle between u = i − 2j − 2kand v = 6i + 3j + 2k.

Example 3:Find the vector projection ofu = 6i + 3j + 2k onto v = i − 2j − 2k,and the scalar component of u in thedirection of v.

Example 4:Determine if the following pairs of vectorare orthogonal:(a) u = 〈3,−2〉 and v = 〈4, 6〉,(b) u = 3i − 2j + k and v = 2j + 4k.


The Cross ProductThe Cross Product of Two Vectors in Space

In studying lines in plane, when we

want to describe how a line was

tilting, we used the notions of slope

and angle of inclination.

In studying lines in space, we want to

describe how a plane is tilting. This

is done by multiplying two vectors in

the plane together to get a third

vector perpendicular to the plane.

The direction of this third vector is

the “inclination” of the plane.

A plane is determined using two

nonzero vectors, u and v, in space

which are not be parallel. A unit

vector n perpendicular to the plane is

selected by the right-hand rule, see

Figure 23.

Figure 23: The construction of u × v.

DEFINITION:

The cross product u × v (“u cross

v”) is the vector

u × v = (|u||v| sin θ)n

The cross product is a vector!noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 29 / 57

The Cross ProductProperties of the Cross Product

Nonzero vectors u and v are parallel if and only if u × v = 0.

Let u, v, w be vectors and r, s be scalars.

P1: (ru) × (sv) = (rs)(u × v)

P2: u × (v + w) = u × v + u × w

P3: v × u = −(u × v)

P4: (v + w) × u = v × u + w × u

P5: 0 × v = 0

P6: u × (v × w) = (u · w)v − (u · v)wFigure 24: The construction of v × u;the unit vector we choose is thenegative, −n, of the one we choose informing u × v.

Applying the definition and P3 to calculate cross products of i, j, and k, we find

i × j = −(j × i) = k

j × k = −(k × j) = i

k × i = −(i × k) = j

i × i = j × j = k × k = 0

Cross product multiplication is not associative so u v w does not generallynoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 30 / 57

The Cross ProductArea of a Parallelogram

Because n is a unit vector, the

magnitude of u × v is

|u × v| = |u||v|| sin θ||n| = |u||v| sin θ

This is the area of the parallelogram

determined by u and v, see

Figure 25, |u| being the base of the

parallelogram and |v|| sin θ| the

height.Figure 25: Area of a parallelogram.


The Cross ProductDeterminant Formula for u × v

Suppose that

u = u1i + u2j + u3k and v = v1i + v2j + v3k

The distributive laws and the rules for multiplying i, j, and k tell us that

u × v = (u1i + u2j + u3k) × (v1i + v2j + v3k)

= u1v1i × i + u1v2i × j + u1v3i × k

+ u2v1j × i + u2v2j × j + u2v3j × k

+ u3v1k × i + u3v2k × j + u3v3k × k

= (u2v3 − u3v2)i − (u1v3 − u3v1)j − (u1v2 − u2v1)k

The component terms in the last line are terms in the expansion of a determinantand, hence, we could calculate the cross product as a determinant

u × v =

˛

˛

˛

˛

˛

˛

i j ku1 u2 u3

v1 v2 v3

˛

˛

˛

˛

˛

˛

For ease in calculating the cross product using determinants, we usually write

vectors in the form v = v1i + v2j + v3k rather than as ordered triples v = 〈v1, v2, v3〉.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 32 / 57

The Cross ProductTorque

We turn a bolt by applying a force F

to a wrench to produce a torque that

causes the bolt to rotate and point in

the direction of the axis of the bolt

according to the right-hand rule, see

Figure 26.

The number we use to measure thetorque’s magnitude is the product ofthe length of the lever arm r and thescalar component of F perpendicularto r, i.e.

Magnitude of torque vector

= |r||F| sin θ

= |r × F|

Figure 26: The torque vector.

If we let n be a unit vector along theaxis of the bolt in the direction of thetorque, then a complete descriptionof the torque vector is r × F, or

Torque vector = (|r||F| sin θ)nnoone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 33 / 57

The Cross ProductTriple Scalar or Box Product

The product (u × v) · w is called the

triple scalar product of u, v, and w

(in that order!!), whose absolute

value

|(u × v) · w| = |u × v||w|| cos θ|

is the volume of the parallelepiped

(parallelogram-sided box)

determined by u, v, and w, see

Figure 27.

|u × v| is the area of the base

parallelogram.

|w|| cos θ| is the height of the

parallelepiped.

Figure 27: Volume of a parallelepiped.

Because of the geometry shown in

Figure 27, (u × v) · w is also called

the box product of u, v, and w.


The Cross ProductTriple Scalar or Box Product

The triple scalar product can be evaluated as a determinant:

(u × v) · w =

»˛

˛

˛

˛

u2 u3

v2 v3

˛

˛

˛

˛

i −

˛

˛

˛

˛

u1 u3

v1 v3

˛

˛

˛

˛

j +

˛

˛

˛

˛

u1 u2

v1 v2

˛

˛

˛

˛

k

–

· w

= w1

˛

˛

˛

˛

u2 u3

v2 v3

˛

˛

˛

˛

− w2

˛

˛

˛

˛

u1 u3

v1 v3

˛

˛

˛

˛

+ w3

˛

˛

˛

˛

u1 u2

v1 v2

˛

˛

˛

˛

=

˛

˛

˛

˛

˛

˛

u1 u2 u3

v1 v2 v3

w1 w2 w3

˛

˛

˛

˛

˛

˛

Hence

(u × v) · w =

˛

˛

˛

˛

˛

˛

u1 u2 u3

v1 v2 v3

w1 w2 w3

˛

˛

˛

˛

˛

˛

(3)


The Cross ProductExamples

Example 1:Find a vector perpendicular to the planeof P(1,−1, 0), Q(2, 1,−1), andR(−1, 1, 2) (Figure 28).

Figure 28: Examples 1 & 2.

Example 2:Find the area of the triangle with verticesP(1,−1, 0), Q(2, 1,−1), and R(−1, 1, 2)(Figure 28).

Example 3:Find a unit vector perpendicular to theplane of P(1,−1, 0), Q(2, 1,−1), andR(−1, 1, 2).

Example 4:Find the volume of the box(parallelepiped) determined byu = i + 2j − k, v = −2i + 3k, andw = 7j − 4k.


Lines and Planes in SpaceLines and Line Segments in Space

This section shows how to use scalar

and vector products to write

equations for lines, line segments,

and planes in space.

In a plane, a line is determined by a

point and the slope of the line.

In space, a line is determined by a

point and a vector giving the line’s

direction.

Figure 29: A line in space.

Suppose that L is a line in space

passing through a point P0(x0, y0, z0

parallel to a vector

v = v1i + v2j + v3k

Then L is a set of points P(x, y, z) for

which⇀

P0P is parallel to v, Figure 29.

Thus,

⇀

P0P = tv

for some scalar parameter t. The

value of t depends on the location of

the point P along the line, and the

domain of t is (−∞,∞).noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 37 / 57


The expanded form of⇀

P0P = tv is

(x − x0)i + (y − y0)j + (z − z0)k = t(v1i + v2j + v3k)

and re-written as

xi + yj + zk = (x0i + y0j + z0k) + t(v1i + v2j + v3k) (4)

Vector equation for a line L through P0(x0, y0, z0) parallel to v is

r(t) = r0 + tv, −∞ < t < ∞ (5)

where r is the position vector of a point P(x, y, z) on L, r0 is the position vector of

P0(x0, y0, z0).

Equating corresponding components of the two sides of Eq. (5) yields parametric

equations for a line L through P0(x0, y0, z0) parallel to v

x = x0 + tv1, y = y0 + tv2, z = z0 + tv3, −∞ < t < ∞ (6)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 38 / 57


Example 1:Find parametric equations for the linethrough (−2, 0, 4) parallel tov = 2i + 4j − 2k (Figure 30).


Example 2:Find parametric equations for the linethrough P(−3, 2,−3) and Q(1,−1, 4).

Example 3:Parametrize the line segment joining thepoints P(−3, 2,−3) and Q(1,−1, 4)(Figure 31).


Example 4:A helicopter is to fly directly from ahelipad at the origin in the direction ofthe point (1, 1, 1) at a speed of 60 m/s.What is the position of the helicopter after10 s?noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 39 / 57

Lines and Planes in SpaceDistance from a Point to a Line in Space

Figure 32: Distance from S to the line throughP parallel to v.

In the notation of Figure 32, distance

from a point S to a line through P

parallel to v is

d =

⇀

PS × v

|v| (7)

Example 5:

Find the distance from the point

S(1, 1, 5) to the line

L : x = 1 + t, y = 3 − t, z = 2t.


Lines and Planes in SpaceEquation for a Plane in Space

A plane in space is determined by knowing a

point on the plane and its tilt or orientation.

This tilt is defined by specifying a vector

that is perpendicular or normal to the plane.

Suppose that plane M passes through a

point P0(x0, y0, z0) and is normal to the

nonzero vector n = Ai + Bj + Ck. Then M is

the set of all points P(x, y, z) for which⇀

P0P

is orthogonal to n (Figure 33). Thus, the dot

product n ·⇀

P0P = 0.

Figure 33: The standard equation for aplane in space is defined in terms of avector normal to the plane: a point Plies in the plane through P0 normal to

n if and only if n ·⇀

P0P = 0.


Lines and Planes in SpaceEquation for a Plane in Space

The plane through P0(x0, y0, z0) normal to n = Ai + Bj + Ck has

Vector Equation: n ·⇀

P0P = 0

Component Equation: A(x − x0) + B(y − y0) + C(z − z0) = 0

Component Equation Simplified: Ax + By + Cz = D

where D = Ax0 + By0 + Cz0

Example 6:

Find an equation for the plane through P0(−3, 0, 7) perpendicular to

n = 5i + 5j − k.

Example 7:

Find an equation for the plane through A(0, 0, 1), B(2, 0, 0), and C(0, 3, 0).


Lines and Planes in SpaceLines of Intersection

Just as lines are parallel if and only if they have the same direction, two planes are

parallel if and only if their normals are parallel, or

n1 = kn2

for some scalar k. Two planes that are not parallel intersect in a line.

Example 8:

Find a vector parallel to the line of

intersection of the planes

3x − 6y − 2z = 15 and 2x + y − 2z = 5,

see Figure 34.

Example 9:

Find parametric equations for the line in

which the planes 3x − 6y − 2z = 15 and

2x + y − 2z = 5 intersect.Figure 34: How line of intersection oftwo planes is related to the planes’normal vectors.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 43 / 57

Lines and Planes in SpaceLines of Intersection

Sometimes we want to know where a line and a plane intersect. For example, if we

are looking at a flat plate and a line segment passes through it, we may be

interested in knowing what portion of the line segment is hidden from our view by

the plate. This application is used in computer graphics.

Example 10:

Find the point where the line

x =8

3+ 2t, y = −2t, z = 1 + t

intersects the plane 3x + 2y + 6z = 6.

Example 10a: Hidden lines in computer graphics

Your eye is at (4, 0, 0). You are looking at a triangular plate whose vertices are at

(1, 0, 1), (1, 1, 0), and (−2, 2, 2). The line segment from (1, 0, 0) to (0, 2, 2) passes

through the plate. What portion of the line segment is hidden from your view by

the plate?

(Hint: This is an exercise in finding intersections of lines and planes.)noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 44 / 57

Lines and Planes in SpaceDistance 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 n = Ai + Bj + Ck is normal to the

plane.

Example 11:

Find the distance from S(1, 1, 3) to the

plane 3x + 2y + 6z = 6, see Figure 35.

Figure 35: The distance from S to the plane

is the length of the vector projection of⇀

PSonto n.


Lines and Planes in SpaceAngles Between Planes

The angle between two intersecting

planes is defined to be the acute angle

between their normal vectors, see

Figure 36.

Example 12:

Find the angle between the planes

3x − 6y − 2z = 15 and 2x + y − 2z = 5.

Figure 36: The angle between two planes isobtained from the angle between theirnormals.


Cylinders and Quadric SurfacesCylinders

Quadric surfaces are surfaces defined by second-degree equations in x, y, and z.

Spheres are quadric surfaces, but there are also other quadric surfaces of equal

interest which will be covered later.

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

(Figure 37).

In solid geometry, where cylinder

means circular cylinder, the

generating curves are circles, but

now we allow generating curves of

any kind.

Figure 37: A cylinder and generating curve.


Cylinders and Quadric SurfacesCylinders

Example 1:

Find an equation for the cylinder

made by the lines parallel to the

z-axis that pass through the parabola

y = x2, z = 0 (Figure 38).

Figure 38: Every point of the cylinder here hascoordinates of the form (x0, x2

0, z).

As Example 1 suggests, any curve f(x, y) = c in xy-plane defines a cylinder parallel

to z-axis whose equation is also f(x, y) = c, e.g. equation x2 + y2 = 1 defines

circular cylinder made by lines parallel to z-axis that pass through circle

x2 + y2 = 1 in xy-plane.

Similarly, any curve g(x, z) = c in xz-plane defines a cylinder parallel to y-axis

whose space equation is also g(x, z) = c. Any curve h(y, z) = c defines a cylinder

parallel to x-axis whose space equation is also h(y, z) = c.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 48 / 57

Cylinders and Quadric SurfacesQuadric Surfaces

We start by studying simple quadric surfaces given by the equation

Ax2 + By

2 + Cz2 + Dz = E

where A, B, C, D, and E are constants.

We shall look at some basic quadric surfaces from this equation which include

ellipsoids (spheres are special cases of ellipsoids),paraboloids,elliptical cones, andhyperboloids.


Cylinders and Quadric SurfacesQuadric Surfaces: ellipsoid

Figure 39: The ellipsoid.

Example 2: The ellipsoid

x2

a2+

y2

b2+

z2

c2= 1

shown Figure 39 cuts the coordinate axes at (±a, 0, 0), (0,±b, 0), and (0, 0,±c). It

lies within the rectangular box defined by the inequalities |x| ≤ a, |y| ≤ b and

|z| ≤ c. The surface is symmetric with respect to each of the coordinate planes

because each variable in the defining equation is squared.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 50 / 57

Cylinders and Quadric SurfacesQuadric Surfaces: ellipsoid

The curves in which the three coordinate planes cut the surface are ellipses. For

example,

x2

a2+

y2

b2= 1 when z = 0.

The curve cut from the surface by the plane z = z0 , |z0| < c, is the ellipse

x2

a2(1 − z0/c)2+

y2

b2(1 − z0/c)2= 1 when z = 0.

If any two of the semiaxes a, b, and c are equal, the surface is an ellipsoid of

revolution. If all three are equal, the surface is a sphere.


Cylinders and Quadric SurfacesQuadric Surfaces: elliptical paraboloid & cone

Figure 40: The elliptical paraboloid.Figure 41: The elliptical cone.


Cylinders and Quadric SurfacesQuadric Surfaces: hyperbolic paraboloid

Figure 42: The hyperbolic paraboloid.

Example 3: The hyperbolic paraboloid

y2

b2− x2

a2=

z

c, c > 0

shown Figure 42, has symmetry with respect to the planes x = 0 and y = 0. The

cross-sections in these planes are:

(a) x = 0 : the parabola z =c

b2y

2, (b) y = 0 : the parabola z = − c

a2x

2.

The parabola in the plane x = 0 opens upward from the origin.The parabola in the plane y = 0 opens downward.noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 53 / 57

Cylinders and Quadric SurfacesQuadric Surfaces: hyperbolic paraboloid

If the surface is cut by a plane z = z0 > 0, the cross-section is a hyperbola,

y2

b2− x2

a2=

z0

c

with its focal axis parallel to the y-axis and its vertices on the parabola in Equation

(1).


Cylinders and Quadric SurfacesQuadric Surfaces: hyperboloid of one sheet

Figure 43: The hyperboloid of one sheet.


Cylinders and Quadric SurfacesQuadric Surfaces: hyperboloid of two sheets

Figure 44: The hyperboloid of two sheets.


Bibliography

1 PETER V. O’NEIL (2012): Advanced Engineering Mathematics, 7ed,ISBN-13: 978-1-111-42741-2, Cengage Learning

2 DANIELA FLEISCH (2012): A Student’s Guide to Vectors and Tensors,ISBN: 978-0-521-17190-8, Cambridge University Press

3 ERWIN KREYSZIG (2011): Advanced Engineering Mathematics, 10ed,ISBN: 978-0-470-45836-5, John Wiley & Sons

4 ALAN JEFFREY (2002): Advanced Engineering Mathematics,ISBN: 0-12-382592-X, Harcourt/Academic Press

5 GLYNN JAMES ET AL. (2011): Advanced Modern Engineering Mathematics, 4ed,ISBN: 978-0-273-71923-6, Pearson Education

6 L. BRIGGS ET AL. (2013): Calculus for Scientists and Engineers: Early Transcendentals,ISBN-13: 978-0-321-78537-4, Pearson Education


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