more vectors. linear combination of vectors or these two vectors are on the same line (collinear)
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More Vectors
Linear Combination of Vectors
Recall that two non-zero vectors and are if, and onlcollinear , for somy if e u v u kv k R
3, 4u
9,12v 1
3u v
or
3v u
These two vectors are on the same line (collinear)
Linear Combination
3,4,Find and such that and are collinea7 ( , 7, ) r.a b u v a b
Since u and v are collinear: 3,4, 7 , 7,k a b
Therefore:
3 1
4 7 2
7 3
ka
k
kb
From (2): 4
7k
Thus: 21 49
,4 4
a b
Linear Combination
Definition: any two non-collinear vectors form a basis for the plane in which they lie, and any other vector in that plane can be written as a linear combination of these basis vectors
(-1,3) (2,4Show that and form a basis for the plane) .u v
We must demonstrate that these two vectors are not collinear 1,3 2,4
u kv
k
Assume
From the 1st element:1
2k
From the 2nd element:3
4k
Since these are different, u and v are not collinear and hence form a basis for the plane.
Linear CombinationWrite as a linear combina(5,8) (-1,3)tion of and (2,4).w u v
w uk lv
5,8 1,3 2,4k l
2 ,3 4k l k l
5 2
8
(1)
(2)3 4
k l
k l
2 (1) 2 2 5
2
5
k
k
23
10l
2 23
5 10w u v
Therefore:
Linear CombinationThe vectors , , and are not coplanar.
Write
(1, 1, 1) (1, -1, 1) (1, 1, -1)
(1 as a linear combination of , , and ,3,1a) )
u v w
x u v w
k lx v mu w
1,3,1 1,1,1 1, 1,1 1,1 1k l m
1,3,1 , ,k l m k l m k l m
(1)1
3 )
(31
(2
)
k l m
k l m
k l m
(1) (3) (2 42 )2k l
(2) (3) 5)2 (k
1
0
l
m
2 0
2
x u v w
x u v
Note: x is coplanar with u and vTherefore:
Thus we have
Linear Combination
For vectors , , and . if and only if, , ( ) 0 , and are coplanar u v w u v w u v w
Theorem:
Are u=(2, -1, -2), v=(1, 1, 1) and w=(1, -5, -7) coplanar?
2, 1, 2 1,1,1 1, 5, 7u v w
1, 4,3 1, 5, 7u v w
1 1 4 5 3 7u v w
0u v w
Therefore the vectors are coplanar
Equations of Lines in the Plane
In order to determine a straight line it is enough to specify either of the following sets of information:
a) Two points on the line, orb) One point on the line and its direction
For a line a fixed vector is called a direction vector for the line if it is parallel to
ll
d
Note: every line has an infinite number of direction vectors that can be represented as where is one direction vector for the line and t is a non-zero real number.
d
td
Equations of Lines in a Plane
Find a direction vector for each linea) The line l1 through points A(4,-5) and B(3, -7)b) The line l2 with slope 4/5
3 4, 7 5 1, 2AThe vector B ��������������
Therefore a direction vector for l1 can be given by vector (-1, -2)
Any scalar multiple of (-1,-2) could also be used as a direction vector of l1a)
b) A line with slope 4/5 that passes through the origin would pass through the point (5,4). Thus we can use direction vector (5,4) for l2
Equations of Lines in the Plane 0Develop a vector equation for the line through the point with direction vector P 3,5 1,2d
Pick any point P(x,y) on the line.
Because P is on the line, the vector P0P (from PO to P, can be written as a scalar multiple of the direction vector d=(1,2): that is
3, 5 1, for any real number2OP P
x t t
d
y
t
��������������
0 0 1 2
through with direction vector is for any real num
A Vec
ber
tor Equation of the Li
, ,
e
,
n
O OP d P P td t
xor y x y t d d
��������������
Equations of Lines (2D)
For each real value of the scalar t in the vector equations corresponds to a point on the line. This scalar is called the parameter for the equation of the line.
0, 0 1 2
0 1 0 2
through with direction vector
Parametric Equ
P ,
ations of the Lin
re
e
aO x y d d d
x x td y y td
d1 and d2 are called direction numbers of the line
Understanding
Find vector and parametric equations of the line through points A(1,7) and B(4,0)
A direction vector for this line is: 4 1,0 7 3, 7d AB ��������������
Thus, a vector equation of this line is: 1,7 3, 7OP t ��������������
From the vector equation we can obtain the parametric equation
1 3
7 7
x t
y t
Equations of Lines (2D)
Yet another form of equation of a line evolves from solving the parametric equations for the parameter.
00 1
1
x xx x td t
d
0
0 22
y yy y td t
d
Therefore
0 0 1 2
0 0
1 2
through with direction
A Symmetric Equation of the Line
vector , , isx y d d
x x y y
d d
Understanding
For each pair of equations, determine whether or not they describe the same line
1, 6 3, 2
4, 4 6,4
) x y r
x s
a
y
2 5
35
)
1
x t y tb
yx
5, 3 2,1
5 3
)
2 1
x y s
x y
c
For each pair of equations, determine whether or not they describe the same line
1, 6 3, 2
4, 4 6,4
) x y r
x s
a
y
Understanding
Step 1: compare the direction vector in both lines 2 3, 2 6,4
The direction vectors are parallel
Step 2: see if a point on one line is also on the other line
Pick the point (4,4) and check
1, 6 3, 2
14 , 6 3, 2
3, 2 3,
4
2
x r
r
r
y
When r=1, we have a match Therefore these equations are for the same line
Understanding
For each pair of equations, determine whether or not they describe the same line
2 5
35
)
1
x t y tb
yx
Step 1: compare the direction vector in both lines
Step 2: see if a point on one line is also on the other line
1 2 11,5 , 1, 5d d d
The direction vectors are parallel
Pick the point (2,0) and check
13
51
30
51
2
15
yx
Since the left-side does not match the right side, the lines are different
Understanding
For each pair of equations, determine whether or not they describe the same line
Step 1: compare the direction vector in both lines 1 22,1 2,1d d
The direction vectors are not parallel, therefore these lines cannot be identical
5, 3 2,1
5 3
)
2 1
x y s
x y
c
Equations of Lines in Space
Vector Equation:
0
A vector Equation through with direction
vector is for any real number
P
O
d OP P td t ��������������
Parametric Equations: 0
0 1
0 2
0
0 0 1 2
3
3through with direction vector , , d= , , areO
x x td
y y td
z z t
d
d
P x y z d d
Symmetric Equations: 0
0
0 0 1 2
0
3
0
1 2 3
through with direction vector are, , d= , ,O
x x
P x y z d d d
y y z z
d d d
Find the vector, parametric, and symmetric equations for the line through the points A(1, 7, -3) and B(4, 0, 2).
First determine the direction vector: 4 1,0 7,2 3 3, 7,5d AB ��������������
A vector equation is: 1,7, 3 3, 7,5OP t ��������������
A parametric equation is: 1 3
7 7
3 5
x t
y t
z t
A symmetric equation is: 1 7 3
3 7 5
x y z
Understanding
Direction NumbersOne alternative technique for describing the direction of a line focusses on the direction angles of the line.
The direction angles of a line in the plane are the angles,
and , , , between a direction vector in the upper
half-plane (where ) and the positive and axi
0
0 s.
l
l
y x y
Direction Numbers in a SpaceThe direction angles of a line in space are the angles, and
, , between a direction vector in the upper
half-space (where ) and the positive and and axe
, ,
0 , ,
0 s.
l
l
z x y z
Direction Cosines
In the Plane: 1 2 2For a line with directi , 0on vector , l d d d d
1cosd
d 2cos
d
d
1 2 3 3For a line with direction vector , , , 0l d d d d d
1cosd
d 2cos
d
d 3cos
d
d
In Space:
Note: 1 2 3
1cos ,cos ,cos , ,
dd d d
d d
The direction cosines of a line are the components of a unit vector in the direction of the line
Understanding
The line l has direction vector (1,3,5). Find its direction cosines and thus its direction angles
1,3,5Let . Thend
2 2 21 3 5
35
d
So, 1cos
1
35
d
d
80
2cos
3
35
d
d
60
3cos
5
35
d
d
32
Understanding
Determine the angle, to the nearest degree, that (1, 2, -3) makes with the positive x-axis.
2 22
cos2 3
1
14
1
1
74.49
75
1cosd
d
UnderstandingFind the cosines for the line: 5
1 2
3
x t
y t
z t
The direction vector is: 1,2, 1d
This vector is not in the upper half-space (last coordinate is not positive)
So we choose: 1, 2,1d
This direction vector (parallel to the first) is in the upper half-space 2 2 2
1 2 1 6d
1cos
6
2cos
6
1cos
6
Understanding
1 3 20,3,2
1A li
4 14ne through the point has direction cosines , , and .
Find parametric equations of
the
l1
e4
in .
l
For the line we could use the direction vector cos ,cos ,cos
1 3 2, ,
14 14 14
d
However, we can obtain “nicer numbers” is we use
1 3 214 , ,
14 14 14
1, 3,2
d
Then a vector equation is: 0,3,2 1, 3,2OP t ��������������
Parametric equations are: 3 3
2 2
x t
y t
z t
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