PERTEMUAN XXI
Definition of Scalar Product
Given vectors A and B as illustrated in Fig. 1, the scalar, or dot product,
between the two vectors is defined as
A • B = | A | | B| cos θ
where
is the angle between the two vectors.
Figure 1 Illustration for definition of dot product.
A • B = B • A is commutative
( A + B ) • C = A • C + B • C distributive.
A • C is the projection of A onto C times the magnitude of C, |C|, and B C is
the projection of B onto C times |C|. Because projections are additive, (11)
follows.
i• i = j• j = k • k = 1
i• j = i• z = j • k = 0
With A and B expressed in terms of these components, it follows from the
distributive and commutative properties that
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A . B = AxBx + AyBy + AzBz
A = A 2 = Ax2 + Ay2 + Az2
Sum and Scalar Multiplication of Vectors
v + w = (a + c,b + d )
and
kv = ( ka, kb )
Example
3 ( 2,1 ) – 2(-1,3) = (6 + 2,3 – 3 )
= ( 8, 0 ) = 8i
Secara geometri v + w merupakan diagonal dari jajaran genjang
dengan sisi v dan w.
v - w = v + (-w).
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Properties of Vector Addition and Subtraction
1.
2.
3.
4.
(u + v) + w = u + (v + w)
a(u + v) = au + av
a(bv) = (ab)v
u+v = v+u
Applications
An boat captain wants to travel due south at 40 knots. If the current is
moving northwest at 16 knots, in what direction and magnitude should he
work the engine?
Solution
u=v+w
where u corresponds to the velocity vector of the boat, v corresponds to
the
engine's vector, and w corresponds to the velocity of the
current. We have
u = -40j
and
w = -8
i+8
j
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Hence
v = u - w = -40j - (-8
i+8
j) = 8
i - (40+8
)j
The magnitude is
[(8
)2 + (40+8
)2]1/2 = 52.5
The direction is
C =A+B=B+A
http://www.phys.hawaii.edu/~teb/java/ntnujava/vector/vector.html
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http://www.pa.uky.edu/~phy211/VecArith/index.html
Dot Product
Definition
We define the dot product of two vectors
v = ai + bj and w = ci + dj
to be
v . w = ac + bd
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Dot Product in R3
If
v = ai + bj + ck and w = di + ej + fk
then
v . w = ad + be + cf
Usaha ( work ) dari F sepanjang PQ :
W = F . PQ
Example
Find the work done against gravity to move a 10 kg baby from the point
(2,3) to the point (5,7)?
Solution
F = ma = (10)(-9.8j) = -98j
v = (5 - 2) i + (7 - 3) j = 3i + 4j
W = F . v = (-98j) . (3i + 4j)
= (0)(3) + (-98)(4) = -392 Tandas minus karena usaha terhadap
grafitasi.
Jadi usaha adalah 392 J .
Definition of Vector Product
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Cross-product dari dua vector A dan B ialah C yang mempunyai besaran
C = A B sinθ
C=A×B
Dot and Cross Product
Visualizing Vectors in 2-D and 3-D
Cross Product
(A+B)×D=A×D+B×D
A×B = - B×A
i x j = - (j x i) = k
j x k = - (k x j) = i
k x i = - (i x k) = j
i
A × B = Ax
Bx
Preview
ixi=jxj=kxk=0
j
Ay
By
k
Az
Bz
Preview
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The Scalar Triple Product
A . ( B × C ) = B . ( C × A ) = C . (A × B )
The Double Cross-Product
The Area of a Parallelogram (1)
The Area of a Parallelogram (2)
Inner Products
Division Vector
Line and Dot product
Distributive Law for Alternate Product
Commutative Law for Inner Product
Unit Vector
Vector Equation of Line
Vector Equation of Line in 3D
Problem of Dot(Inner) Product
Sum of Vectors
Linearly Independent of Vector
Equation of Plane
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Example
The dot product of a = [ 1, 3, -2 ] and b = [ -2, 4, -1 ] is
akan memberikan theta = 45.6 derajat.
Example [ 1, -1, 3 ] dan [ 3, 3, 0 ] orthogonal karena dot product = 0
1 (3) + ( -1 ) ( 3 ) + 3 ( 0 ) = 0.
Example
Cross product dari a = [3, -2, -2 ] dan b = [ -1, 0, 5 ] adalah

Multiplication by scalars:
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


Distributivity:
The vector triple product of the vectors a, b, and c:
The scalar triple product of the vectors a, b, and c:
Volume Paralelepipedum
A.(BxC)
Luas Jajaran Genjang
| A × B|
Example
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Cari luas segitiga yang titik sudutnya adalah (1,1,3), (4,-1,1), dan
(0,1,8).
Jawab :
Bentuk vektor a dari (1,1,3) ke (4,-1,1) dan b dari (1,1,3) ke
(0,1,8).
yaitu a = ( 2 -1, 2-(-1), 2-3 ) = (3,-2,-2) dan b = (-1, 0, 5 ). Maka
luas jajaran genjang adalah :
Sehingga luas segitiga adalah 8.26.
EXAMPLE
Given A = 3i + 2j - k and B = i - 3j + 4k, find A + B and A - 3B.
Answer
A + B = 4i - j + 3k
A - 3B = 11j - 13k
EXAMPLE Find the vector from P1(3, 2, -1) to P2(-4, 6, 5).
Answer
P1P2 = -7i + 4j + 6k
EXAMPLE Find the vector 3 units long in the direction of
A = i + j + k.
Answer
First, we must find the length of A.
Now to find the directions of A.
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Now to find the new vector with length of 3 units in the
directions of A.
EXAMPLE
Given P (2, 4, 5) and Q (3, -1, 3), find (a) PQ, (b) the
distance between P and Q, (c) the direction of PQ, and (d)
the midpoint of the line segment PQ.
SOLUTION:
a. PQ = (3 - 2)i + (-1 - 4)j + (3 - 5)k = i - 5j - 2k
b. The distance between P and Q is the length of the
vector PQ.
c. The direction of PQ.
d. The midpoint of the line segment PQ is
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DOT PRODUCTS
EXAMPLE Given A = 2i + 3j and B = i - j + 2k, find (a) A  B, (b) | A |, |
B |, (c) the cosine of the angle between A and B, and (d) find
 to the nearest hundredth of a radian.
Answer
A  B = -1
EXAMPLE Given A = i - j + k and B = 3i + 2j - k, find (a) A  B, (b) | A
|, | B |, (c) the cosine of the angle between A and B, and (d)
find  to the nearest hundredth of a radian.
Answer
a) 0
b) √3
c) √14
d) π/2
VECTOR PROJECTIONS
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The vector projection of B = PQ onto a
nonzero vector A = PS is the vector PR
determined by dropping a perpendicular
from Q to the line PS. (see figure 2)
The notation for this vector is Proj A B.
figure 2
| B | cos  is called the scalar component of B in the direction of A.
EXAMPLE
Given A = -i + j and B = 2i + j - k, find (a) A  B, | A |, | B |, (b) the
cosine of the angle between A and B, (c) the scalar component of B
in the direction of A, and (d) the vector projection, Proj A B.
Answer
A  B = (-1)(2) + (1)(1) + (0)(-1) = -2 + 1 = -1
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EXAMPLE Find the vector projection, Proj A B when A = 5j - 3k
and
B = i + j + k.
Answer
A  B = (0)(1) + (5)(1) + (-3)(1) = 5 - 3 = 2
A  A = (5)(5) + (-3)(-3) = 25 + 9 = 34
Work
The work done by a constant force F along PQ is given by
W = F . PQ
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Work = (scalar component of F in the direction of D)*(length of D)
= (| F | cos  ) | D| = F  D.
Example
Cari usaha yang diperlukan untuk mengangkat benda 10 kg dari titik ( 2,3 )
ke titik ( 5,7)
Solution
Dari fisika diperoleh
F = ma = (10)(-9.8j) = -98j
v = (5 - 2) i + (7 - 3) j = 3i + 4j
Usaha adalah
.
.
W = F v = (-98j) (3i + 4j)
= (0)(3) + (-98)(4) = -392
EXAMPLE
Find the work done by a force F = 5k (magnitude 5 N) in
moving an object along a line from the origin to the point
(1, 1, 1). (Distance is in meters.)
SOLUTION: We need to find the displacement D first. It is the vector
from the origin to the point (1, 1, 1).
D=i+j+k
Work = F  D = (0)(1) + (0)(1) + (5)(1) = 5 N - m = 5 J
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EXAMPLE
Find the angle between the lines 3x - 2y = 4 and x - 4y = 8.
Answer
First put each equation into the slope-intercept form.
Now determine the related vectors.
v1 = 2i + 3j
v2 = 4i + j
v1  v2 = (2)(4) + (3)(1) = 8 + 3 = 11
EXAMPLE
Find the angles between the curves y = x 3 and x = y 2. (There will
be two intersection points.)
Answer
First, find the points of intersection of the two curves.
y = (y 2) 3  y = y 6  y 6 - y = 0  y (y 5 - 1) = 0  y =
0 or y = 1 When y = 0, then x = 0, and when y = 1, then x
= 1.
Now find the derivative of each function.
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(0, 0) m1 = 0
v1 is a horizontal
v1 = i
vector.
m2 is
v2 is a vertical vector.
v2 = j
undefined
v1  v2 = 0  cos  = 0   =  /2
Each point of intersection will have two angles, if  = 
/2, then the other angle  =  -  /2 =  /2.
(1, 1) m1 = 3
m2 = 1/2
v1 = i + 3j
v2 = 2i + j
v1  v2 = (1)(2) + (3)(1) = 2 + 3 = 5
EXAMPLE
Given P (1, 1, 1), Q (2, 1, 3), and R (3, -1, 1), find a) the
area of the triangle determined by the points P, Q, and R,
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and b) the unit vector perpendicular to the plane PQR.
Answer
First of all, I must construct two vectors out of the three
points. I will find the vectors PQ and PR and these two
vectors lie in the plane that contains the points P, Q, and R.
PQ = (2 - 1)i + (1 - 1)j + (3 - 1)k = i + 2k
PR = (3 - 1)i + (-1 - 1)j + (1 - 1)k = 2i - 2j
a. If the area of the parallelogram defined by the
vectors A and B is
| A x B |, then if I take half of this area, I will have
the area of the triangle.
Area of the triangle = (1/2) | A x B |
Area of the triangle = (1/2)(6) = 3
b. The unit vector.
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EXAMPLE
Find the area of the parallelogram with vertices A(-6, 0),
B(1, -4), C(3, 1), and D(-4, 5).
Answer
EXAMPLE
Find the area of the triangle with vertices A (2, 4), B (3, 4),
and C (6, 0).
Answer
Area of the triangle ABC = 4/2 = 2.
TRIPLE SCALAR PRODUCT or BOX PRODUCT
FACT:
EXAMPLE
Find (A x B)  C and the volume of the parallelepiped
Answer
defined by A = i + j - 2k, B = -i - k, and C = 2i + 4j - 2k.
Volume of the parallelepiped | A x B |  C = | 8 | = 8
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THE DISTANCE FROM A POINT TO A LINE IN SPACE
To find the distance from a point S
to a line that passes through the
point P parallel to a vector v, we
will find the length of the
component of PS normal to the line.
(see figure 1)
figure 1
EXAMPLE
Find the distance from (-1, 4, 3) to the line x = 10 + 4t, y = -3,
SOLUTION: z = 4t.
Therefore v = 4i + 4k.
To find a point on the line, let P be of the form (x 0, y 0, z 0) or
(10, -3, 0).
Now find the vector PS where S is (-1, 4, 3).
PS = (-1 - 10)i + (4 - (-3))j + (3 - 0)k = -11i + 7j + 3k
The vector parallel to the line is of the form Ai + Bj + Ck.
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EQUATIONS FOR PLANES IN SPACE
Suppose plane M passes through a point
P0 (x 0, y 0, z 0) and is normal to the
nonzero vector n = Ai + Bj + Ck. (see
figure 2) Then M is the set of all points
P (x, y, z) for which P0P is orthogonal to
n. (i.e. P lies on M if and only if n  P0P
= 0) We will use the equation
figure 2
n  P0P = 0.
FACT: Ai + Bj + Ck is normal to the plane Ax + By + Cz = 0.
EXAMPLE
Find the equation of the plane through (2, 4, -1) parallel to the
plane 2x - y + 3z = 5.
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SOLUTION: First of all, we must determine the vector n that is normal to the
plane
2x - y + 3z = 5.
n = 2i - j + 3k
Since the new plane is to be parallel to the plane 2x - y + 3z = 5,
then n will also be normal to the new plane. Now to find the
vector P0P where P is (x, y, z).
P0P = (x - 2)i + (y - 4)j + (z + 1)k.
Now to use n  P0P = 0.
(x - 2)(2) + (y - 4)(-1) + (z + 1)(3) = 0
2x - 4 - y + 4 + 3z + 3 = 0
2x - y + 3z = -3
EXAMPLE
Find the equation of the plane through (1, 1, 1), (2, -1, 0), and (3,
1, 5).
SOLUTION:
8x + 6y - 4z = 10
EXAMPLE
Find the point of intersection of the lines x = t, y = -t + 2, z = t + 1,
and x = 2s + 2, y = s + 3, z = 5s + 6, and then find the plane
determined by these lines.
SOLUTION: Line 1: x = t, y = -t + 2, z = t + 1
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Line 2: x = 2s + 2, y = s + 3, z = 5s + 6
I need to find the point of intersection of these two lines. To do it, I
will solve the equations simultaneously.
EXAMPLE
Find the equation of the line through (2, -1, 1) perpendicular to
the plane 3x - 2y + 6z = 5.
SOLUTION:
x = 2 + 3t , y = -1 - 2t , z = 1 + 6t
EXAMPLE
Find the distance from the point S (1, 0, -1) to the plane -4x + y +
z = 4.
SOLUTION:
Let use find a point on the plane. A point that is easy to find is
one of the three intercepts. I will find the y-intercept and that is
when x = z = 0. Therefore, y = 4.
Let P (0, 4, 0). (The y-intercept.)
So PS = (1 - 0)i + (0 - 4)j + (-1 - 0)k = i - 4j - k
n = -4i + j + k
I need the direction of n because I am going to use the following
formula.
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ANGLES BETWEEN PLANES
The angle between two
intersecting planes is defined to be
the (acute) angle determined by
their normal vectors. (see figure 3)
figure 3
SOLUTION:
Find the angles between the planes 5x + y - z = 10 and x 2y + 3z = -1.
SOLUTION:
For plane 1: (5x + y - z =
n1 = 5i + j - k
10)
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For plane 2: (x - 2y + 3z = -
n2 = i - 2j + 3k
1)
n1  n2 = (5)(1) + (1)(-2) + (-1)(3) = 0
EXAMPLE
A baseball is thrown from the stands, 32 ft above the field at
an angle 30o up from the horizontal. When and how far away
will the ball strike the ground if its initial speed is 32 ft/sec?
(see figure 2)
SOLUTION: Gravity for this problem is
32.
figure 2
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EXAMPLE 2:
In Moscow in 1987, Natalya Lisouskaya set a women's world
record by putting an 8-lb 13-oz shot 73 ft 10 in. Assuming
that she launched the shot at a 40 o angle to the horizontal 6.5
ft above the ground, what was the shots initial speed. (see
figure 3)
SOLUTION: 73 ft 10 in  73.83 ft
g = 32
x = v0 cos 40 o t
y = 6.5 + (v0 sin 40 o)t - 16t 2
I will solve for t by setting x =
73.83.
figure 3
Plug this value into y. The position of the shot put at that
value of t will be zero.
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EXAMPLE 3:
What two angles of elevation will enable a projectile to reach
a target 16 km down range on the same level as the gun if the
projectile's initial speed is 400m/sec? (See figure 4)
SOLUTION: g = 9.8
16 km = 16000 m
x = (400 cos  ) t
y = (400 sin  ) t - 4.9t 2
I will set x = 16000 and
figure 4
solve for t.
When t is the above expression, y will be zero. I will solve
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16000tan  - 7840sec 2  = 0
16000tan  - 7840(tan 2  + 1) = 0
-7840tan 2  + 16000tan  - 7840 = 0
TERIMA KASIH
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