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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
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 3 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 4 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 5 / 57
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〉
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 10 / 57
VectorsVector Algebra Operations
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
VectorsVector Algebra Operations
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
VectorsVector Algebra Operations
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 14 / 57
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
«
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 17 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 18 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 20 / 57
The Dot ProductAngle Between Vectors
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
The Dot ProductAngle Between Vectors
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 23 / 57
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 Dot ProductDot Product Properties and Vector Projections
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
The Dot ProductDot Product Properties and Vector Projections
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 28 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 31 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 34 / 57
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)
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 35 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 36 / 57
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
Lines and Planes in SpaceLines and Line Segments in Space
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
Lines and Planes in SpaceLines and Line Segments in Space
Example 1:Find parametric equations for the linethrough (−2, 0, 4) parallel tov = 2i + 4j − 2k (Figure 30).
Figure 30: Example 1.
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).
Figure 31: Example 3.
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 40 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 41 / 57
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).
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 42 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 45 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 46 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 47 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 49 / 57
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.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 51 / 57
Cylinders and Quadric SurfacesQuadric Surfaces: elliptical paraboloid & cone
Figure 40: The elliptical paraboloid.Figure 41: The elliptical cone.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 52 / 57
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).
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 54 / 57
Cylinders and Quadric SurfacesQuadric Surfaces: hyperboloid of one sheet
Figure 43: The hyperboloid of one sheet.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 55 / 57
Cylinders and Quadric SurfacesQuadric Surfaces: hyperboloid of two sheets
Figure 44: The hyperboloid of two sheets.
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 56 / 57
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
noone�dev.null MKMM 1213 Advanced Engineering Mathematics Vectors & Geometry of Spaces 57 / 57