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PART I: APPLICATION OF VECTORS
AND VECTOR-VALUED FUNCTIONS
Lines in Space
Vectors that connect any pair of points on a line l are k.
Vector description of a line
Consider the locus of the terminal points of the vectors (first in 2D)
~
tA
−∞<t<∞
It is a straight line. This line can be displaced to
~
describe any straight line k to A.
The general vector expression for a line in space is:
~
~r = ~r0 + tA
−∞<t<∞
~ is a
where ~r0 is an arbitrary point on the line and A
vector k to the line.
Example Find a vector equation of the line that contains (−1, 3, 0) and is parallel to 2¯i − 3¯j − k¯
Example Find a vector equation of the line that
passes through the two points: (1, 2, 3) and
(4, 5, 6).
Note ~r(t) gives the location of a point on the line. It
~ is. Namely, if P
~
is not parallel to the line. A
~ are any two different points on the line,
and Q
~
then P~Q k A.
Parametric equations
x = x0 + tA1
y = y0 + tA2
z = z0 + tA3
Example Find a set of parametric equations for the
line of the previous example.
Analytic geometry (AG) description of a line
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
Example Find an AG description of the line in the
previous example.
Degrees of Freedom / Dimensions
≡ number of free parameters used to describe the
geometrical object
~ be a vector in space, the object tA
~
Example Let A
(−∞ < t < ∞) has one free parameter,
therefore the object has dimension 1.
~ and B
~ are not parallel, and both 6= ~0,
Example If A
what is the object
~ + sB
~ −∞ < t < ∞ −∞ < s < ∞ ?
tA
Planes in 3D Space
Basic geometrical facts
(i) Two different intersecting lines l1 , l2 (directions given
~ B
~ respectively) define a plane through them.
by A,
(ii) There is a unique line l3 that passes through the
point of intersection and is ⊥ to both lines. A vector
~ k l3 defines a normal to the plane that contains
N
~ ·A
~=N
~ ·B
~ = 0)
l1 , l2 . (N
(iii) There is only one plane that contains a given point
and is ⊥ to a given nonzero vector.
(i) can be viewed as the definition of a plane; (ii) & (iii) come from the three-dimensional nature of space.
Vector description of a plane
A plane through two intersecting lines l1 and l2 can
be generated as:
~ + sB
~
~r = ~r0 + tA
− ∞ < t < ∞, −∞ < s < ∞
~ B
~
where ~r0 is the point of intersection of l1 , l2 , and A,
are vectors along l1 , l2 respectively.
A normal is perpendicular to all directed line seg~ · (tA
~ + sB)
~ = 0.
ments on the plane as N
Recall (ii) above for the definition of a normal. A vector is a normal to a plane iff it is perpendicular to all
vectors associated with the directed line segments on the plane.
Parametric description of a plane:
Decompose the vector equation into components, as
(linear) functions of the parameters.
x = x0 + tA1 + sB1
y = y0 + tA2 + sB2
z = z0 + tA3 + sB3
This approach is not used much for planes, but very
useful for general surfaces. The coordinates of a
point on a surface can be specified as:
x = f1 (t, s),
y = f2 (t, s),
z = f3 (t, s).
These are two-variable functions which make components of a function
f : R2 → R3 .
AG description of a plane:
Let P be a fixed point on the plane, Q be any point
~ be a normal. Then
on the plane, and N
~ · P~Q = 0
N
~ = (x, y, z), and N
~ =
Writing P~ = (x0 , y0 , z0 ), Q
(a, b, c), one has
(a, b, c)· [(x, y, z) − (x0 , y0 , z0 )] =
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.
The resultant equation is of the form
ax + by + cz = d.
A normal to the plane can be found easily as
~ = a~i + b~j + c~k.
N
Example: Find an equation of the plane that contains
the point (−2, 4, 5) and has normal vector
7~i − 6~k
Example: Find a unit normal to the plane
x + y + z = 1.
Example: Find the plane through the three points
(0, 0, 1), (1, 0, 1), (0, 1, 1).
Quadric Surfaces
Definition: A quadric surface is a surface which contains points that satisfies the 2nd degree
polynomial equation (AG description)
Ax2 +By 2 +Cz 2 +Dxy+Eyz+F zx+Gx+Hy+Iz+J = 0
where A, ..., J are constants.
Quadric surfaces fall into nine major classes. Examples
of each are provided as following (a, b, c > 0):
1.
2.
3.
4.
5.
6.
7.
8.
9.
x2
y2
z2
Ellipsoid
+ 2 + 2 =1
2
a
b
c
x2
y2
z2
Hyperboloid of one sheet
+ 2 − 2 =1
a2
b
c
y2
z2
x2
− 2 − 2 =1
Hyperboloid of two sheets
2
a
b
c
x2
y2
z2
Elliptic double cone
+ 2 − 2 =0
2
a
b
c
x2
y2
z
Elliptic paraboloid
+ 2 =
a2
b
c
x2
y2
z
Hyperbolic paraboloid
−
=
a2
b2
c
y2
x2
+ 2 =1
Elliptic cylinder
a2
b
x2
y2
Hyperbolic cylinder
− 2 =1
2
a
b
z
Parabolic cylinder x2 =
c
—– Problem Set 1 —–
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