Math 283
Section 9.1 Notes
3-D Coordinate Systems
Example:
Give a geometric description of the set of points in space whose coordinates satisfy the
given pairs of equations.
a. x = -1, z = 0
b. x 2 + y 2 =
4, z =
−3
25, y =
−4
c. x 2 + y 2 + z 2 =
Example:
Describe the sets of points in space whose coordinates satisfy the given inequality or
combinations of equations and inequalities.
1.
a. 0 ≤ x ≤ 1
b. 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
c. 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1
2.
0
a. x 2 + y 2 ≤ 1, z =
b. x 2 + y 2 ≤ 1, z =
3
c. x 2 + y 2 ≤ 1, no restriction on z
Example:
Describe the given set with a single equation or with a pair of equations.
The plane through the point (3, -1, 2) perpendicular to the
a. x- axis
b. y-axis
c. z-axis
Example:
Find the distance between points P1 and P2.
a. P1(-1, 1, 5) and P2(2, 5, 0)
b. P1(5, 3, -2) and P2(0, 0, 0)
Example:
Find the center and radii of the spheres.
x2 + y 2 + z 2 − 6 y + 8z =
0
Math 283
Section 9.2 Notes
Vectors
Definition:
1. A vector is a directed line segment.
2. 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 the same length and direction.
Notation:
2-D Vector




v v1i + v2 j
v = v1 , v2 or =
3-D Vector





v = v1 , v2 , v3 or v = v1i + v2 j + v3 k
Magnitude or Length:

v =
v12 + v2 2 or

v =
v12 + v2 2 + v32 or
( x2 − x1 ) + ( y2 − y1 )
2
2
2-D
( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 )
2
Unit Vector:

 v
u= 
v
Vector made from length and direction

  v
v=v 
v
Midpoint:
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
2
2
3-D
Example:
Given:

=
u 3, −2

v = −2,5
Find the component form and magnitude of the vector.

a. −2v
 
b. −2u + 5v
Example:
Find P1 P2 if P1 (1, 2, 0 ) and P2 ( −3, 0,5 ) . Write in i, j, k component form.
Example:
Express the vector
 as a product of its length and direction.



a. v = 9i − 2 j + 6k
b. Given: P1 (1, 4,5 ) and P2 ( 4, −2, 7 ) and find the midpoint.
Math 283
Section 9.3 Notes
The Dot Product


The Dot Product: Let u = u1 , u2 , u3 and v = v1 , v2 , v3 .
u ⋅ v= u1v1 + u2 v2 + u3v3
The angle between two nonzero vectors.
 
u ⋅v
u v
θ=  


Vectors u and v are orthogonal (or perpendicular) if and only if u ⋅ v =
0 .
Vector projection of u onto v:
 u ⋅v 
projv u =

v
 v2 


Scalar component of u in the direction of v:
u cos θ =
u ⋅v
v
= u⋅
v
v
Example:
Given:




v =2i + 10 j − 11k

 

u = 2i + 2 j + k
   
a. Find v ⋅ u , v , u
b. Find the angle between the vectors.
c. Find the scalar component of vector u in the direction of vector v.
d. Find the vector projv u
Example:
Given:
 

v =−i + j




u = 2i + 3 j + 2k
   
a. Find v ⋅ u , v , u
b. Find the angle between the vectors.
c. Find the scalar component of vector u in the direction of vector v.
d. Find the vector projv u
Application Example:
A water main is to be constructed with a 20% grade in the north direction and a 10%
grade in the east direction. Determine the angle θ required in the water main for the
turn from north to east.
Math 283
Section 9.4 Notes
The Cross Product
Definition:
u×v =
( u v sin θ ) n where n is the unit (normal) vector
Parallel Vectors:
  
Nonzero vectors u and v are parallel if and only if u × v =
0
Area of a Parallelogram
=
u × v u v=
sin θ n u v sin θ
Example:
 
 
Find the length and direction of u × v and v × u


 


a. u =2i + 3 j and v =−i + j
 3 1  
   
b. u = i − j + k and v = i + j + k
2
2
Example:
Given: P(1, 1, 1), Q(2, 1, 3), and R(3, -1, 1)
a. Find the area of the triangle determined by the points P, Q, and R.
b. Find a unit vector perpendicular to the plane PQR.
Example:
Find the area of the parallelogram whose vertices are given:
A(0, 0), B(7, 3), C(9, 8), D(2, 5)
Math 283
Section 9.5 Notes
Lines and Planes in Space
Vector Equation for a Line
A vector equation for the line L through P0(x0, y0, z0) parallel to v is

r ( t )= r0 + tv ,
−∞ < t < ∞
Where r is the position vector of a point P(x, y, z) on L and r0 is the position vector of
P0(x0, y0, z0).
Parametric Equations for a Line
The standard parametrization of the line through P0(x0, y0, z0) parallel to v =v1i + v2 j + v3 k
is
x= x0 + tv1 , y= y0 + tv2 , z= z0 + tv3, − ∞ < t < ∞
Distance from a Point S to a Line through P Parallel to v
d=
PS × v
v
Equation for a Plane
The plane through P0(x0, y0, z0) normal to n = Ai + Bj + Ck has
0
Vector equation: n ⋅ P0 P =
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:
Find the parametric equations for the lines.
a. The line through P(1, 2, -1) and Q(-1, 0, 1)
b. The line through the point (3, -2, 1) and parallel to the line x = 1 + 2t, y = 2 – t, z = 3t.


 
 
c. The line through (2, 3, 0) perpendicular to the vector u =i + 2 j + 3k and v =3i + 4 j + 5k
Example:
Find the equations for the plane.
a. The plane through (1, -1, 3) parallel to the plane 3x + y + z = 7.
b. The plane through (2, 4, 5), (1, 5, 7), and (-1, 6, 8).
Example:
Find the plane determined by the intersecting lines.
L1 : x = t , y = 3 − 3t , z = −2 − t , − ∞ < t < ∞
L2 : x = 1 + s, y = 4 + s, z = −1 + s, − ∞ < s < ∞
Example:
Find a plane through the points P(1, 2, 3) and Q(3, 2, 1) and perpendicular to the plane
4x – y + 2z = 7.
Example:
Find the distance from the point to the line.
(0, 0, 0); x = 5 + 3t, y = 5 + 4t, z = -3 – 5t
Example:
Find the distance from the point to the plane.
(2, 2, 2), 2x + y + 2z = 4
Example:
Find the angle between the planes.
5x + y – z = 10 and x – 2y + 3z = -1
Example:
Find the point in which the line meets the plane.
x = 2, y = 3 + 2t, z = -2 – 2t
6x + 3y – 4z = -12
Example:
Find the parametrization for the line in the planes.
3x – 6y – 2z = 3 and 2x + y – 2z = 2
Math 283
Section 9.6
Cylinders and Quadric Surfaces
Types of Surfaces:
x2 y 2 z 2
1. Ellipsoid: 2 + 2 + 2 =
1
a
b
c
x2 y 2 z
2. Elliptical Paraboloid: 2 + 2 =
a
b
c
2
2
2
x
y
z
3. Elliptical Cone: 2 + 2 =
a
b
c2
x2 y 2 z 2
4. Hyperboloid of One Sheet: 2 + 2 − 2 =
1
a
b
c
z 2 x2 y 2
5. Hyperboloid of Two Sheets: 2 − 2 − 2 =
1
c
a
b
y 2 x2 z
6. Hyperboloid Paraboloid: 2 − 2 =
, c>0
b
a
c
Example:
Graph
a. =
z y2 −1
36
b. 4 x 2 + y 2 =
16
c. 4 x 2 + 4 y 2 + z 2 =
d. z =8 − x 2 − y 2
9 y2
e. 4 x 2 + 9 z 2 =
1
f. y 2 + z 2 − x 2 =