Write the point-slope form of a line.. y = y 1 + b (x - x 1 )
v Initial point terminal point A(x 1, y 1 ) B(x 2, y 2 ) V =AB, vector from the point A to the point...
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Transcript of v Initial point terminal point A(x 1, y 1 ) B(x 2, y 2 ) V =AB, vector from the point A to the point...
v
Initial point
terminal point
A(x1 , y1)
B(x2 , y2)
V=AB, vector from the point A to the point B,
is a directed line segment between A and B.
v
A(x1 , y1)
B(x2 , y2)
MAGNITUDE OF VECTOR
║V║ = (x2−x1)2 + (y2−y1)
2
U
UNIT VECTOR
║U║ = 1
MAGNITUDE OF VECTOR IS 1
EXAMPLES
1- Vector v has the initial point A(3, 4) and the terminal point B( -2, 5).
Magnitude of vector v is (-2-3)2 + (5-4)2 = (-5)2 + 12 = 25+1 = 26
║V║= 26
* V is not a unit vector
2- Vector i has the initial point (0, 0) and the terminal point(1, 0).
Magnitude of the vector i is
= (1-0)2 + (0-0)2 = 12 + 02 = 1+ 0 = 1
3- Vector j has the initial point (0, 0) and the terminal point (0 ,1).
Magnitude of the vector j is
= (0-0)2 + (1-0)2 = 12 + 02 = 1+ 0 = 1
* i is a unit vector
* j is a unit vector
║ i ║
║j║
SCALAR MULTIPLICATION OF VECTOR
V
W−W
2W
0.5V
−2W
k
4k
VECTOR ADDITION
V
W
W+V
TRIANGLE METHOD
VECTOR ADDITION
V
W
W+V
PARALLELOGRAM METHOD
V
W
W+V
-W
V-W
VECTOR ADDITION
-V
W-V
VECTORS IN A COORDINATE PLANE
x
y
A(x1 , y1)
B(x2 , y2)
C(x2−x1 , y2−y1)
O
W=AB V
Vectors W and V are equivalent vectors. Since V starts from the origin we use a special notation for V
V = x2−x1 , y2−y1
VECTORS IN A COORDINATE PLANE
x
y
A(5 , 1)
B(1 , 4)C(−4 , 3)
O
W = AB
V
Vectors W and V are equivalent vectors.
V = −4 , 3
*−4 is the x-component ,and 3 is the y-component of the vector V
3
4
1
5-4 1
VECTORS IN A COORDINATE PLANE
x
y
A(1 , 4)
B(-3 , 1)
C(-4 , -3)
W = AB
V
Vectors W and V are
equivalent vectors. V = -4 , -3
- 3
4
1
- 3-4
1
VECTORS IN A COORDINATE PLANE
x
y
i
Vectors i and j are special unit vectors.
j = 0 , 1 1
1
j
i = 1 , 0
VECTORS IN A COORDINATE PLANE
Find a vector that has the initial point (3 , -1) and is equivalent to V = -2 , 3 .
x
A(3 , -1)
P(-2 , 3)
B( 1 ,2)
W = AB
Vectors W and V are equivalent vectors.
V
- 1
3
1- 2
2
3W
If ( x, y) is the terminal point of W, then
x−3 = −2 → x = 1 and
y−(−1) = 3 → y = 2
y
BASIC VECTOR OPERATIONS
V = a , b and W = c , d are two vectors and k is a real number.
1- ║V║ = a2 + b2
2- v+w = a , b + c , d = a+c , b+d
3- kV =k a , b = ka , kb
4- ║kV║ = k ║V║
BASIC VECTOR OPERATIONS
5V = 5 -2 , 3 = -10 , 15
║v║ = (-2)2 + 32 = 4 + 9 = 13
V +W = -2 , 3 + 4 , -1 = -2+4, 3-1 = 2 , 2
-3W = -3 4 , -1 = -12 , 3
V = -2 , 3 , W = 4 , -1
5V −3W = -10 , 15 + -12 , 3 = -10-12 , 15+3 = -22 , 18
ANY VECTOR CAN BE WRITTEN IN TERMS OF THE UNIT VECTORS i AND j
If V = a , b is any vector, then by using basic vector operations we get ;
V = a , b = a , 0 + 0 , b
= a 1 , 0 + b 0 , 1 = ai+ bj
V = a , b = ai + bj
i
j
4i
3j
4i+3j
P(4 , 3)
x
y
V = 4i+3j = 4 , 3
VECTORS WRITTEN IN TERMS OF THE UNIT VECTORS i AND j
VECTORS WRITTEN IN TERMS OF THE UNIT VECTORS i AND j
i j
4i
-3j
4i-3j
P(4 , -3)
x
y
V = 4i-3j = 4 , -3
DIRECTION ANGLE OF VECTORS
x
y
α direction angle of V β
V
W
V = x , y = V cosα , sinα =
P(x , y)
V cosα , V sinα
Q(a , b)
W = a , b = W cosβ , sinβ = W cosβ , W sinβ
Vcosα =
x
sinα = V
y
x
ytan
DIRECTION ANGLE OF VECTORS
If V = -2i + 3j , then find the direction angle of V.
tanα = = = −yx
3-2
V = -2i+3j = -2 , 3
α = tan-1[− ] in the second quadrant
32
32
DIRECTION ANGLE OF VECTORS
If V = , − , then find the direction angle of V.
tanα = = − = −yx
α = tan-1[− ] = −
3
12
32
3
123
2
3π
DIRECTION ANGLE OF VECTORS
If ║V║ = 6 and the direction angle of V i s , then
find the x and y-components of V. 67π
V = x , y =
V cosα , V sinα
V = = 6cos , 6sin6
7π6
7π−6 , −6
23 1
2
−3 3 , −3V=
UNIT VECTORS ON THE SAME DIRECTION WITH A GIVEN VECTOR
If V = x , y , then
U = , = V
x
V
y
is the unit vector on the same direction with V.
V = x , y = V
cosθ , sinθ
U
θ is the direction angle of V
UNIT VECTORS ON THE SAME DIRECTION WITH A GIVEN VECTOR
If V = -3 , 4 , then find the unit vector on the same direction.
U = , = V
x
V
y
v = (-3)2 + 42 = 9 + 16 = 25 = 5
5-3
54,
Now find the vector W on the same direction with magnitude 6.
W = 6U = 6 =
See the illustrations on the next slide
5-3
54,
5-18
524,
x
y
P(-3 , 4)
O
U
V
4
-3
W
V = 5U
W = 6U
X = -3U
X
║W║= 6
║V║ = 5
║X║=3
If V = −2i + 4j , then find the vector on the opposite direction
with magnitude 6.
First, find the unit vector on the direction of V.
U = = ,
Now, multiply the unit vector with −6. That will give you the answer.
W = −6U
The vector on the opposite direction with magnitude 6.
V
x
V
y −220
420
EXAMPLE
DOT PRODUCT OF VECTORS
V = a , b and W = c , d
V∙W = ac + bd
V∙V = a2 + b2 = ║V║2
DOT PRODUCT OF VECTORS
1− V∙W = W∙V dot product is commutative 2− U∙(V + W) = U∙V + U∙W distributive 3− a (V∙W) = (aV )∙W=V∙(aW) ,a is a scalar 4− V∙V = ║V║2
5− 0∙W = 0 zero vector
6− i∙i = j∙j = 1
7− i∙j = j∙i = 0
DOT PRODUCT OF VECTORS
║U+V║2 = (U+V).(U+V) = U.U+U.V+V.U+V.V
║U+V║2 = ║U║2 + 2U.V + ║V║2
SIMILARLY
║U−V║2 = (U−V).(U−V) = U.U−U.V−V.U+V.V
║U−V║2 = ║U║2 − 2U.V + ║V║2
ANGLE BETWEEN TWO VECTORS
VW
θ angle between V and W θ
V∙W = ║V║║W║cos θ
cos θ = __________║V║║W║
V∙W
EXAMPLE
V = −2 , 3 , W = 4 , − 1
V∙W = ac + bd = −2.4 + 3.(− 1) = − 8 − 3 = −11
V∙V = a2 + b2 = (−2)2 + 32 = 13 = ║V║2
cos θ = __________ =
_______║V║║W║
V∙W −1113 17
EXAMPLE
V = 3 , −6 , W = −1 , 2
V∙W = 3. (−1) − 6.2 = − 3−12 = −15
║V║= 45 , ║W║= 5
cos θ = __________ =
_______ =
______ =
_____ = −1║V║║W║
V∙W −1545 5
−15225
−1515
cos θ = −1, then θ = cos-1(−1) = π
EXAMPLE
V
W
If V∙W = 0 ,then V and W are perpendicular
θ is 90◦ , cos θ = 0
θ
EXAMPLE
If V = 4i-3j, then find a vector that is perpendicular to V
V∙W = 4x – 3y = 0
If W = xi + yj, then
Any choice of x and y that satisfies the equation above is an answer
Since 3 and 4 satisfy the equation
W = 3, 4 is one of the vectors
V
W
V and W are parallel
V
W
θ is 0◦, cos θ = 1
θ is 180◦, cos θ = -1
Same direction
Opposite direction
EXAMPLE
V = 3 , −6 , W = −1 , 2
cos θ = __________ =
_______ =
______ =
_____ = −1║V║║W║
V∙W −1545 5
−15225
−1515
cos θ = −1, then θ = cos-1(−1) = π
V and W are parallel with opposite direction
EXAMPLE
V = 3 , −6 , W = 1 , −2
cos θ = __________ =
_______ =
______ =
___ = 1║V║║W║
V∙W 1545 5
15225
15 15
cos θ = 1, then θ = cos-1(1) = 0
V and W are parallel with same direction
EXAMPLE
V = 3 , −6 , W = 4 , 2
cos θ = __________ =
_______ = 0║V║║W║
V∙W 045 20
cos θ = 0 , then θ = cos-1(0) = 90◦
V and W are perpendicular ( orthogonal ) vectors
EXAMPLE
If ║U + V║ = 7 , U is a unit vector and cos θ = ____ where θ is the
angle between U and V, then find the magnitude of V.
12
║U+V║2 = ║U║2 + 2U.V + ║V║2 = 7
1 + 2U.V + ║V║2 = 7 , ║V║2 + 2U.V − 6 = 0
cos θ = __________ =
___ , 2U.V = ║V║║U║║V║
U∙V 1 2
║V║2 + 2║V║ − 6 = 0 (║V║−2)(║V║+3) = 0
║V║= 2 or ║V║= −3, ║V║≥ 0 so ║V║= 2
SCALAR PROJECTION
VW
θ
projW
V = ║V║cos θ = _______
║W║
V∙W
EXAMPLE
If V = 2i + 2j , and W = −4i−2j , then find projW
V
projW
V = _______ = _____ = ______ = ____
║W║V∙W -12
20 -12 2 5
-6 5
EXAMPLE
If V = kW for any nonzero number k , then
projW
V = _______ = _________ = _________ = ________
║W║V∙W (kW)∙W
║W║k(W∙W)
║W║k║W║
2
║W║
projW
V = k║W║
How about projV
W ?.