chapter 3 : vectorschapter 6 : vectors 6.1 lines in space ... 6.2 planes in space 6.2.1 intersection...
TRANSCRIPT
1
CHAPTER 6 : VECTORS
6.1 Lines in Space
6.1.1 Angle between Two Lines
6.1.2 Intersection of Two lines
6.1.3 Shortest Distance from a Point to a Line
6.2 Planes in Space
6.2.1 Intersection of Two Planes
6.2.2 Angle between Two Planes
6.2.3 Angle between a Line and a Plane
6.2.4 Shortest Distance from a Point to a Plane
Review:
Basic Concepts
Vectors in Space
The Dot Products
The Cross Products
2
Basic Concepts
What is scalar?
a quantity that has only magnitude
What is vector?
a quantity that has magnitude and
direction
A vector can be represented by a directed line
segment where
length of the line segment
- the magnitude of the vector
direction of the line segment
- the direction of the vector
3
2 2,Q x y
Terminal Point
1 1,P x y
Initial Point
PQ
A vector can be written as PQ , or a . The
order of the letters is important. PQ
means the vector is from P to Q or the
position vector Q relative to P, QP means
vector is from Q to P or the position vector
P relative to Q.
If P 11, yx is the initial point and
Q 22 , yx is the terminal point of a directed
line segment, PQ then component form
of vector v that represents PQ is
4
12
12
12121 ,,yy
xxyyxxvv x
and the magnitude or the length of v is
2 2
2 1 2 1
2 2
1 2
x x y y
c c
v
1 1,P x y
2 2,Q x y
1 1 1 10, 0 ,OP x y x y
2 2 2 20, 0 ,OQ x y x y
-Note-
Any vector that has magnitude of 1 unit = unit
vector.
5
Example:
Find the component form and length of the
vector v that has initial point 7,3 and
terminal point 5,2 .
Solution:
2 3,5 7 5,12v
2 2
5 12 25 144 13v
Example:
Given 5,2v and 4,3w , find each of
the following vectors:
a) v2
1
b) vw c) wv 2
Answer:
a) 1,5 2 b) 5, 1 c) 4,13
6
Theorem:
If a is a non-null vector and if a is a unit
vector having the same direction as a, then
ˆ
aa =
a
-Note-
To verify that magnitude is 1, a = 1
Example:
Find a unit vector in the direction of
5,2v and verify that it has length 1.
Solution:
2 2
2,5 12,5
292 5
v
2 2
2 29 5 29 1 1v
7
Standard Unit Vectors
Three standard unit vectors are: i, j dan k
z
y
x
1
1
1
i
j
k
Vectors i, j and k can be written in
components form:
i = < 1, 0, 0 >,
j = < 0, 1, 0 > and
k = < 0, 0, 1>
and can interpreted as
a = <x, y, z>
= xi + yj + zk
8
The vector QP with initial point
111 ,, zyxP and terminal point 222 ,, zyxQ
has the standard representation
kji )()()( 121212 zzyyxxQP
or
121212 ,, zzyyxxPQ
Example:
Let u be the vector with initial point (2,-5)
and terminal point (-1,3), and let jiv 2 .
Write each of the following vectors as a linear
combination of i and j.
a) u
b) vuw 32
9
0x
y
v
v
1x
1y
-Note-
If is the angle between v and the positive
x – axis then we can write
cos and y sinx v v ;2 2
1 2v x y .
Example:
The vector v has a length of 3 and makes an
angle of 630
with the positive x-axis.
Write v as a linear combination of the unit
vectors i and j.
10
Vectors in Space
Properties of Vectors in Space
Let 321 vvv ,,v and 321 www ,,w be
vectors in 3 dimensional space and k is a
constant.
1. wv if and only if
332211 wvwvwv ,, .
2. The magnitude of v is 2
3
2
2
2
1 vvv v
3. The unit vector in the direction of v is
v
,,
v
v 321 vvv
4. 332211 wvwvwv ,,wv
5. 321 kvkvkvk ,,v
6. Zero vector is denoted as 0 000 ,, .
7. + v w w v
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8. Let 1 2 3, ,u u uu ,
then + + u + v w u v w .
9. + u 0 u
10. +u -u = 0
11. +dc +d u = cu u
12. +c c cu + v u v
13. c d cdu u
14. 1 and 0u = u u = 0
15. c =cu u
Example:
Express the vector PQ if it starts at point
),,( 856P and stops at point ),,( 937Q in
components form.
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Solution:
2 1 2 1 2 1, ,PQ x x y y z z
7 6,3 5,9 8PQ
1, 2,1PQ
Example:
Given that 213 ,,a , 461 ,,b . Find
(a) ba 3 (b) b
(c) a unit vector which is in the direction of b.
(d) find the unit vector which has the same
direction as ba 3 .
Answer:
(a) 0,19,10 (b) 53 (c) 41 23
2, ,6 3
(d) 1 (e) 1
1,6,453
(f)
10,19,10
461
13
Parallel Vector
have same slopes
1 2 ; constantsv v
So, there are multiples of each other.
Example:
Vector w has initial point (2,-1,3) and
terminal point (-4,7,5). Which of the
following vectors is parallel to w?
Solution:
4 2,7 1,5 3w
6,8,2w
One example: 2 6,8,2 12,16,4
Another is 1
6,8,2 3,4,12
.
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Example: (Collinear Points)
Determine whether the point 0,1,2,3,2,1 QP
and 6,7,4 R lie on the same line.
Solution:
2 1,1 2,0 3 1,3, 3PQ
22 21 3 3 19PQ
4 2,7 1, 6 0 2,6, 6QR
22 22 6 6 2 19QR
4 1,7 2, 6 3 3,9, 9PR
22 23 9 9 3 19PR
Thus, PR PQ QR . Since one vector is a
multiple of the other, the two vectors are
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parallel and since they share a common point
Q, they must be the same line.
The Dot Product
-Theorem-
If 321 vvv ,,v and 321 www ,,w , then the
scalar product wv is
wv 321 vvv ,, 321 www ,,
332211 wvwvwv
-Note-
The dot product is also called
- the scalar product
- the inner product
The dot product of two vectors is a scalar.
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The Angle between Vectors
Refer to the figure below, let
1 2 3, ,u u OP u u u ,
1 2 3, ,v v OQ v v v
be two vectors and let be the angle between
them, with 0 .
z
y
x
1 2 3, ,Q v v v 1 2 3, ,P u u u
vu
c
Compute the distance, c between points P and
Q in two ways.
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1) Using the Distance formula
2 2 22
1 1 2 2 3 3
2 2 2 2 2 2
1 2 3 1 2 3
1 1 2 2 3 32
c u v u v u v
u u u v v v
u v u v u v
2 2
1 1 2 2 3 32u v u v u v u v ---(1)
2) Using the Law of Cosines
2 22 2 cosc u v u v ---(2)
Equating equation (1) and (2), we get
1 1 2 2 3 3cos
u v u v u v u v
u v u v
Example:
If v = 2i-j+k, w = i+j+2k and the angle
between v and w is 60, find wv .
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Solution:
22 2 2 2 22 1 1 1 1 2 cos 3v w
6 6 cos 3 6 1 2 3
Example:
Given that 3,2,2 u , 1,8,5v and
2,3,4 w , find
(a) vu
(b) wvu
(c) wvu
(d) the angle between u and v
(e) the angle between v and w.
Answer:
(a) -3
(b) 12, 9,6
(c) -23
19
(d) 94 24
(e) 87 46
Example:
Let A=(4,1,2), B=(3,4,5) and C=(5,3,1) are
the vertices of a triangle. Find the angle at
vertex A.
Answer:
79 12
-Theorem-
The nature of an angle , between two vectors
u and v.
is an acute angle if and only if 0 vu
is an obtuse angle if and only if 0 vu
= 90 if and only if 0 vu . The
Vectors u and v are orthogonal /
perpendicular.
20
Example:
Show that the given vectors are perpendicular
to each other.
(a) i and j
(b) 3i-7j+2k and 10i+4j-k
-Theorem-
(Properties of Dot Product)
If u,v and w are nonzero vectors and k is a
scalar,
1. uvvu
2. wuvuwvu
3. vuvu kk
4. 2
vvv
5. 0 u00u
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The Cross Products
- The cross product (vector product) vu is
a vector perpendicular to u and v.
(illustrated in figure below)
- The direction is determined by the right
hand rule.
u v
u v
If the first two fingers of the right hand
point in the directions of andu v
respectively, then the thumb points in
the direction of u v .
Ex: i j k
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- The length is determined by the lengths of
u and v and the angle between them.
- If we change the order informing the cross
product, then we change the direction.
Ex:
v u u v
-Theorem-
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
23
Properties of Cross Product
(a) 0uu
(b) uvvu
(c) wuvuwvu
(d) vuvuvu kkk
(e) u // v if and only if 0u v
(f) u 0u00
Example:
1) Given that 4,0,3u and 2,5,1 v ,
find
(a) vu
(b) uv
2) Find two unit vectors that are
perpendicular to the vectors u 2i+2j-3k
and v = i+3j+k.
24
Answer:
1) (a) -20i+10j+15k (b) 20i-10j-15k
2)
111, 5,4
162
(The unit vector in the
opposite direction is also a unit vector
perpendicular to both u and v )
Further geometry interpretation of the cross
product comes from computing its magnitude.
2 2 2 2
2 3 3 2 3 1 1 3 1 2 2 1u v u v u v u v u v u v u v
2 2 2 2 2 2 2
1 2 3 1 2 3
2
1 1 2 2 3 3
2 2 2
2 2 2 2 2
2 2 2
2 2 2
cos
1 cos
sin
u v u u u v v v
u v u v u v
u v u v
u v u v
u v
u v
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with is the angle between u and v .
Therefore, sin .u v u v
A D
B C
u
v
sinu
From the figure above, we can see that the
magnitude of the cross product is the area of
the parallelogram of which arrows
representing the two vectors are adjacent
sides.
Area of a parallelogram vuvu sin
Area of triangle = vu2
1
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Example:
(a) Find an area of a parallelogram that is
formed from vectors u = i + j - 3k and
v = -6j + 5k.
(b) Find an area of a triangle that is formed
from vectors u = i + j - 3k and
v = -6j + 5k.
Answer:
(a) 230 (b) 230
2
Scalar Triple Product
-Theorem-
If 111 ,, zyxa , 222 ,, zyxb and
333 ,, zyxc ,
then
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333
222
111
zyx
zyx
zyx
cba
Properties of The Scalar Triple Product
1) cbacba
2) a b c a c b b c a
3) a c b a b c
4) 0 a a b
5) a d b c a b c d b c
Example:
If a = 3i + 4j - k, b = -6j+5k and c = i+j-k,
evaluate
(a) cba (b) cba
(c) bac (d) cab
28
6.1 Lines in Space
In this section we use vectors to study lines in
three-dimensional space.
HOW LINES CAN BE DEFINED USING VECTORS?
The most convenient way to describe a line in
space is to give a point on it and a nonzero
vector parallel to it.
Suppose L is a straight line that passes
through ),,( 000 zyxP and is parallel to the
vector kjiv cba .
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Thus, a point ),,( zyxQ also lies on the line if
vPQ t .
Let,
0r OP and r OQ ,
Then
0r rPQ .
vrr t 0
vrr t 0
cbatzyxzyx ,,,,,, 000
-Theorem-
(Parametric Equations for a Line)
The line through the point ),,( 000 zyxP and
parallel to the nonzero vector cba ,,A has
the parametric equations,
30
atxx 0 , btyy 0 , ctzz 0 .
If we let 000 ,, zyx0R denote the position
vector of ),,( 000 zyxP and zyx ,,R the
position vector of the arbitrary point Q(x,y,z)
on the line, then we write equation (1) in the
vector form,
ARR t 0 .
Example:
Give the parametric equations for the line
through the point (6,4,3) and parallel to the
vector 7,0,2 .
-Theorem-
(Symmetric Equations for a line)
31
The line through the point ),,( 000 zyxP and
parallel to the nonzero vector cba ,,A has
the symmetrical equations,
c
zz
b
yy
a
xx 000
.
Example:
Given that the symmetrical equations of a line
in space is 2
4
4
3
3
12
zyx.
Find,
(a) a point on the line.
(b) a vector that is parallel to the line.
32
6.1.1 Angle between Two Lines
Consider two straight lines
c
zz
b
yy
a
xxl 1111 :
and
f
zz
e
yy
d
xxl 2222 :
The line 1l parallel to the vector kjiu cba
and the line 2l parallel to the vector
kjiv fed . Since the lines 1l and 2l are
parallel to the vectors u and v respectively,
then the angle, between the two lines is
given by
vu
vu cos
33
Example:
Find an acute angle between line
1l = i + 2j + t (2i – j + 2k)
and line
2l = 2i – j + k + s (3i – 6j + 2k).
34
6.1.2 Intersection of Two lines
In three dimensional coordinates (space),
two line can be
in one of the
three cases as
shown below,
35
Let 1l and 2l are given by:
)2(:
(1)and:
2222
1111
f
zz
e
yy
d
xxl
c
zz
b
yy
a
xxl
From (1), we have cba ,,1v
From (2), we have fed ,,2v
Two lines are parallel if we can write
21 vv
The parametric equations of 1l and 2l are:
tczz
tbyy
taxxl
1
1
11 :
)3(
:
2
2
22
fszz
esyy
dsxxl
36
Two lines are intersect if there exist unique
values of t and s such that:
fsztcz
esytby
dsxtax
21
21
21
Substitute the value of t and s in (3) to get x, y
and z. The point of intersection = (x, y, z)
Two lines are skewed if they are neither
parallel nor intersect.
Example:
Determine whether 1l and 2l are parallel,
intersect or skewed.
a) tztytxl 74,41,33:1
szsysxl 73,45,32:2
37
b) zyx
l
4
2
1
1:1
3
23
1
4:2
zy
xl
Solutions:
a) for 1l :
point on the line, P = (3, 1, - 4)
vector that parallel to line, 7,4,31v
for 2l :
point on the line, Q = (2, 5, 3)
vector that parallel to line, 7,4,32v
?21 vv
1where21 vv
Therefore, lines 1l and 2l are parallel.
b) Symmetrical equations of 1l and 2l can be
rewrite as:
38
1
0
4
2
1
1:1
zyxl
3
)2(
1
3
1
4:2
zyxl
Therefore:
for 1l : P = (1, 2, 0) , 1,4,11v
for 2l : Q = (4, 3, -2) , 3,1,12v
?21 vv
21 vv not parallel.
In parametric eq’s:
1
2
: 1 , 2 4 ,
: 4 , 3 , 2 3
l x t y t z t
l x s y s z s
1 4 (1)
2 4 3 (2)
2 3 (3)
t s
t s
t s
39
Solve the simultaneous equations (1), (2), and
(3) to get t and s.
5 7and
4 4s t
The value of t and s must satisfy (1), (2) and
(3). Clearly they are not satisfying (2) i.e
7 5 1 172 3 ?
4 4 4 4
Therefore, lines 1l and 2l are not intersect.
This implies the lines are skewed!
Example:
Let L1 and L2 be the lines
tztytxL 51,45,41:1
tztytxL 5,34,82:2
(a) Are the lines parallel?
(b) Do the lines intersect?
40
6.1.3 Shortest Distance from a Point to a Line
Q
sinQP
P
V
Distance from a point Q to a line that passes
through point P parallel to vector v is equal to
the length of the component of PQ
perpendicular to the line.
sinQPd
where is the angle between v and vector
QP
.
41
Since
sinQPQP
vv ,
so we have the shortest distance of Q from L
as
v
v QPd
Example:
Find the distance from the point (0, 0, 0) to
the line,
5
3
4
5
3
5
zyx.
Example:
Find the distance from the point (2, 1, 3) to
the line,
3,61,22 ztytx
42
Example:
Find the shortest path from the point Q(2, 0, -
2) to the line
2
1
1
1
3
2:
zyxl
6.2 Planes in Space
Suppose that is a plane. Point ),,( 000 zyxP
and ),,( zyxQ lie on it. If kji cbaN is a
non-null vector perpendicular (ortoghonal) to
, then N is perpendicular to PQ .
P 0 0 0( , , )x y z Q ( , , )x y z
43
Thus,
0NPQ
0,,,, 000 cbazzyyxx
0)()()( 000 zzcyybxxa
-Conclusion-
The equation of a plane can be determined if a
point on the plane and a vector orthogonal to
the plane are known.
-Theorem-
(Equation of a Plane)
The plane through the point ),,( 000 zyxP and
with the nonzero normal vector cba ,,N
has the equation
44
Point-normal form:
0000 zzcyybxxa
Standard form:
ax by cz d with 0 0 0d ax by cz
Example:
1. Give an equation for the plane through the
point (2, 3, 4) and perpendicular to the
vector 4,5,6 .
2. Give an equation for the plane through the
point (4, 5, 1) and parallel to the vectors
A= 1,0,2 and B 4,1,0 .
45
3. Give parametric equations for the line
through the point (5, -3, 2) and
perpendicular to the plane 5726 zyx .
6.2.1 Intersection of Two Planes
Intersection of two planes is a line. (L)
To obtain the equation of the intersecting line,
we need
a point on the line L which is given by
solving the equations of the two planes.
a vector parallel to the line L which is
46
21 NN
If cbaNN ,,21 , then the equation of
the line L in symmetrical form is
c
zz
b
yy
a
xx 000
Example:
Find the equation of the line passing through
P(2,3,1) and parallel to the line of intersection
of the planes x + 2y - 3z = 4 and x - 2y + z = 0.
Answer:
2 3 1
4 4 4
x y z
47
6.2.2 Angle between Two Planes
-Properties of Two Planes-
An angle between the crossing planes is an
angle between their normal vectors.
1 2
1 2
cosN N
N N
Two planes are parallel if and only if their
normal vectors are parallel,
1 2N N .
Two planes are orthogonal if and only if
1 2 0N N .
Example:
Find the angle between plane 043 yx and
plane 522 zyx .
48
6.2.3 Angle between a Line and a Plane
Let be the angle between the normal vector
N to a plane and the line L. Then we have
Nv
Nv cos
where v is vector parallel to L. Furthermore, if
the angle between the line L and the plane ,
then
2
2
49
cos2
sinsin
Nv
Nv sin
Example:
Find the angle between the plane
523 zyx and the line
3
3
1
2
2
3
zyx.
50
6.2.4 Shortest Distance from a Point to a
Plane
(a) From a Point to a Plane
-Theorem-
The distance D between a point ),,( 111 zyxQ
and the plane dczbyax is
222
111
N
N
cba
dczbyaxPQD
Where ),,( 000 zyxP is any point on the plane.
),,( 111 zyxQ
),,( 000 zyxP
D
N
D
51
Example:
Find the distance D between the point
(1, -4, -3) and the plane 1632 zyx .
Example:
1) Show that the line
1
1
23
1
zyx
is parallel to the plane 123 zyx .
2) Find the distance from the line to the plane
in part (a).
(b) Between two parallel planes
The distance between two parallel planes
1dczbyax and 2dczbyax is given by
222
21
cba
ddD
52
Example:
Find the distance between two parallel planes
322 zyx and 7442 zyx .
-Note-
Both formulas can also be used to compute
the distance between 2 skewed lines.
(c) Between two skewed lines
N u v
v
2L
u
P
Q
d
1L
53
Assume L1 and L2 are skew lines in space
containing the points P and Q and are parallel
to vectors u and v respectively.
Then the shortest distance between L1 and L2
is the perpendicular distance between the two
lines and its direction is given by a vector
normal to both lines.
So, the distance between the two lines is
absolute value of the scalar projection of PQ
on the normal vector.
cosd
PQ
u v PQN PQ
N u v
54
Example:
Find the shortest distance from P(1, -1, 2) to
the plane 3x – 7y + z = 5.
Example:
Find the shortest distance between the skewed
lines.
l1 : x = 1+2t, y = -1+ t , z = 2 + 4t
l2 : x = -2+4s, y = -3s , z = -1+s
Example:
Find the distance between the lines
)(32:1 kitkjiL
tztyxL 3,21,0:2
55
Example:
Find the distance between the lines L1 through
the points A(1, 0, -1 ) and B(-1, 1, 0) and the
line L2 through the points C(3, 1, -1) and
D(4, 5, -2) .