VECTOR ANALYSIS International Program on Science Education
Faculty of Mathematics and Sciences Education Indonesia University of Education
Vectors and Scalars
A vector is a quantity having both magnitude and direction, direction, such as displacement, velocity, force, and acceleration. A scalar is a quantity having magnitude but no direction, direction, e. g. mass, length, time, temperature, volume, speed and any real number.
Vectors and Scalars Graphically a vector is represented by an arrow OP (Fig.. 1) defining the direction, (Fig the magnitude of the vector being indicated by the length of arrow.. arrow Magnitude of vector is determined by arrow, using precise unit unit..
Symbol and Notation of Vector Vector is denoted by bold face type such as Av or it can be represented by A The magnitude is denoted by r A or A Vector is drawn by arrow. arrow. Tail of arrow show position of Origin or initial point while the head of arrow show terminal point or terminus terminus..
Graphic of Vector P
A
O
Origin point
Figure 1
Terminal point
Definition
Two vector A and B are equal equal,, if they have the same magnitude and direction regardless of the position of their initial points. Thus A = B in Fig. 2 A B Figure 2
Definition
A vector having direction opposite to that of vector A but having the same magnitude is denoted by –A. A=-B B=-A A
B Figure 3
Resultant of Vector Definition: The sum or resultant of vector A and B is a vector C formed by placing the initial point of B on the terminal point of A to the terminal point of B (Figure 4)
Definition
The Sum of Vector C = A + B VECTOR A
B
B C=A+B
A
A
C=A+B Figure 4
B
The difference of vector
The difference of vector A and B, represented by: D = A - B = A + ((-B) VECTOR -B B D=A-B A A
-B D=A-B A
Definition
The product of vector A by scalar m is a vector mA Vector
A
C
If, m = 3 C = 3A
Laws of Vector Algebra 1.
2.
3.
4.
5. 6.
If A, B, C are vectors and m, n are scalars. A+B=B+A Commutative Law for Addition A+(B+C)=(A+B)+C Associative Law for Addition mA=Am Commutative Law for Multiplication m( m(n nA)=mn )=mn((A)=n( )=n(m mA) Associative Law for multiplication (m+n m+n))A =mA+nA +nA Distributive Law m( m(A+B A+B)) =m =mA+mB Distributive Law
A Unit Vector
A Unit Vector is a vector having unit magnitude, if A is a vector with magnitude A≠0, then A/A is a unit vector having the same direction as A. Any vector A can be represented by a unit vector a in the direction of A multiplied by the magnitude of A. In symbols, A = Aa Aa
Components of a Vector A = A1i + A2j + A3k A1i = component of vector A in the direction of xxaxis A2j = component of vector A in the direction of yyaxis A3k = component of vector A in the direction of zzaxis x
z
A A3k k A1i
i
j A2j
y
Addition of Vector A = A1i + A2j + A3k B = B1i + B2j + B3k C = A + B = (A1i + A2j + A3k) + (B1i + B2j + B3k) C = A + B = (A1+B1)i + (A2+B2)j + (A3+B3)k C = A - B = (A1i + A2j + A3k) - (B1i + B2j + B3k) C = A - B = (A1-B1)i + (A2-B2)j + (A3-B3)k
Vector Multiplication with scalar A = A1i + A2j + A3k B = B1i + B2j + A3k
D = 3A = 3(A1i + A2j + A3k)
Magnitude of Vector Phytagoras Teorema :
z
(OP)2 = (OQ)2 + (QP)2
P
but (OQ)2 = (OR)2 + (RQ)2
A A3k
so (OP)2 = (OR)2 + (RQ)2 + (QP)2 or A2
= A1 + A2 + A3 2
2
A=
A12 + A22 + A32
O y
R
2
or
A1i
x
A2j
Q
Example: Known r1= 2i+ i+4 4j-5k and and r2 = i+2j+3 j+3k a. Determine vector resultant r1 and r2 ! b. Determine unit vector in vector resultant direction ! Answer :
a. R = r1+r2 =(2i+ i+4 4j-5k) + (i+ (i+2 2j+3 j+3k) = 3i + 6j – 2k b. R = (3i + 6 j − 2k ) = 9 + 36 + 4 =
R 3i + 6 j − 2 k r = = R 7 Ceck magnitude of unit vector = 1
49 = 7
Solved Problems 1.
(a) (b) (c) (d) (e)
State which of the following are scalars and which are vectors. (scalar) (vector) (f) Energy Weight (scalar) (g) Volume (scalar) Calorie Specific heat (scalar) (h) distance (scalar) (scalar) Momentum (vector) (i) speed (scalar) (j) magnetic field intensity Density (vector)
2. (a)
(b)
Represent graphically: A force of 10 N in a direction 30˚ north of east. A force of 15 N in a direction 30˚ east of north.
Unit = 5 N N
N
15 N 30˚
10 N
30˚
W S
Figure (a)
E
E
W S
Figure (b)
to be continued... Thanks...
Perkalian Titik (Dot Product) Dot poduct antara A dan B Atau perkalian skalar didefinisikan : A . B = AB cos θ θ Adalah sudut terkecil yang diapit A dan B Secara fisis dot product adalah proyeksi suatu vektor terhadap vektor lainnya, sehingga sudut yang diambil adalah sudut yang terkecil
Perkalian Titik (Dot Product) A . B = (A A1i + A2j + A3k).(B1i + B2j + B3k) = (A A1i ).(B1i + B2j + B3k) + (A A2j).(B1i + B2j + B3k) + (A A3k).(B1i + B2j + B3k) = A1B1(i.i) + A1B2(i.j (i.j) + A1B3(i.k (i.k) + A2B1(j.i) + A2B2(j.j) + A2B3(j.k) + A3B1(k.i) + A3B2(k.j) +A3B3(k.k)
A.B = A1B1 + A2B2 + A3B3
i.i = i i cos 0o =1 i. j = j.i = i j cos 90o = 0
j. j = j j cos 0 o =1
j.k = k . j = j k cos 90o = 0
k .k = k k cos 0o =1
i.k = k .i = k i cos 90o = 0
Contoh dot product dalam Fisika F θ
F θ
S
W = FS cos θ =
F
F.S
W = usaha
θ
F = Vektor gaya S
S = Vektor perpindahan
Contoh dot product dalam Fisika nA
nA
B
B θ
θ
φ = BA cos θ =
B.A
φ = Fluks magnetik B = Medan magnetik A = arah bidang
Catatan : Bidang adalah vektor memiliki luas dan arah. Arah bidang adalah arah normal bidang di suatu titik. Normal = tegak lurus
Perkalian Silang (Cross Product) Cross poduct antara A dan B Atau perkalian vektor didefinisikan : A x B = AB sin θ u θ Adalah sudut terkecil yang diapit A dan B Hasil perkalian silang antara vektor A dan vektor B adalah sebuah vektor C yang arahnya tegak lurus bidang yang memuat vektor A dan B, sedemikian rupa sehingga A, B, dan C membentuk sistem tangan kanan (sistem skrup)
Perkalian Silang (Cross Product)
Perkalian Silang (Cross Product) B θ
C B
A
θ
-C A
C=AxB
-C = B x A
Pada sistem koordinat tegak lurus i×i =0
j× j =0
k ×k =0
i× j =k
j ×k =i
k ×i = j
j ×i =− k
k× j =−i
i×k =− j
j
k
i
Perkalian silang (Cross Product) A x B = (A A1i + A2j + A3k) x (B1i + B2j + B3k) = (A A1i )x(B1i + B2j + B3k) + (A A2j)x(B1i + B2j + B3k) + (A A3k)x(B1i + B2j + B3k) = A1B1(ixi) + A1B2(ixj (ixj) + A1B3(ixk (ixk) + A2B1(jxi) + A2B2(jxj) + A2B3(jxk) + A3B1(kxi) + A3B2(kxj) +A3B3(kxk)
= A1B1(0) + A1B2(k) + A1B3(-j) + A2B1(-k) + A2B2(0) + A2B3(i) + A3B1(j) + A3B2(-i) +A3B3(0)
Perkalian silang (Cross Product) A x B = A1B1(0) + A1B2(k) + A1B3(-j)
+ A2B1(-k) + A2B2(0) + A2B3(i) + A3B1(j) + A3B2(-i) +A3B3(0) A x B = (A A1B2 - A2B1) k + (A3B1-A1B3) j + (A2B3 - A3B2) i
i
j
k
A × B = A1 B1
A2 B2
A3 B3
Contoh perkalian silang dalam Fisika F θ r O
r r r r τ = r × F = r F sin θ = r × F
Contoh Soal Jika gaya F = 2i - j + 3k bekerja pada titik (2,-1,1), tentukan torsi dari F terhadap titik asal koordinat
Gerak melingkar ω
v P
θ
r
r r r v =ω×r
Perkalian tiga vektor
(
r r r r r r r r r B C sin θ A cos φ = B × C A cos φ = A • B × C
)
Aplikasi Perkalian Skalar Tiga Vektor F
n r
L
O
Komponen torsi terhadap garis L :
τ II
(
r r = nˆ •τ = nˆ • r × F r
)
Contoh Soal Jika gaya F = i + 3j – k bekerja pada titik (1,1,1), tentukan komponen torsi dari F terhadap garis r = 3i + 2k + (2i - 2j + k)t. )t.
Solusi: Pertama kita tentukan vektor torsi terhadap sebuah titik pada garis yaitu titik (3,0,2). Torsi tersebut adalah τ = r x F dimana r adalah vektor berasal dari titik pada garis ke titik dimana dimana F bekerja, yaitu dari (3,0,2) ke (1,1,1), sehingga r = (1,1,1) - (3,0,2) = (-2,1,-1). Dengan demikian vektor torsi τ :
r r r τ =r × F
Contoh Torsi:
Torsi untuk garis adalah n.( .(rrxF) dimana n adalah vaktor satuan sepajang garis, dengan n = 1/3(2i 1/3(2i2j+k). Kemudian torsi untuk garis adalah n.( .(rrxF) = 1/3(2i 1/3(2i2j+k).(2 ).(2ii-3j-7k)=1
Aplikasi Tripel Scalar Product
Aplikasi Tripel Scalar Product salah satunya pada momentum linear
PERSAMAAN GARIS LURUS DAN PERSAMAAN BIDANG
Persamaan Garis Lurus y (x,y) B
Q (y-y0)
(x-x0)
A
(x0,y0) P
r0
r
garis
b
a
x
Definisi Garis Apakah garis itu? Garis adalah deretan titiktitik-titik secara kontinu Dari gambar : B = r – r0 dan A // B (Perbandingan setiap komponen akan sama dimana B = (xi (xi+yj +yj)-(x0i+y0j) = (x(x-x0)i+(y (y--y0)j dan A = ai ai+bj +bj
sehingga
x − x0 y − y0 → 2D = a b x − x0 y − y0 z − z0 = = → 3D a b c Disebut persamaan persamaan garis lurus simetris (x0,y0,z0) adalah suatu titik yang dilalui garis a,b,c. Komponen vektor arah.
Dari gambar di atas juga :
r = r0 + B dan B = tA sehingga r = r0 +At +At = (x0,y0,z0) + (a,b,c)t atau r = ix ix0 + jy jy0 + kz kz0 + (a (ai+ i+b bj+z j+zk)t k)t Disebut persamaan persamaan garis lurus parametrik
Contoh Tentukan persamaan garis lurus parametrik dan simetrik yang melalui titik (2,1,5) dan titik (3,3,1)! (3,3,1)
(2,1,5) x0=2 y0=1 zo=5
A
Solusi A = (3,3,1) – (2,1,5) = (1,2,(1,2,-4) A = i+2j +2j-4k a = 1, 1, b = 2, 2, c = -4 Sehingga : r = (2,1,5) + (1,2,(1,2,-4)t atau r = 2i+j+5k +5k+(i +(i+2j +2j-4k)t Titik yang dilalui
Arah garis
Persamaan garis parametrik
Lanjutan… x − x0 y − y0 z − z0 = = a b c x − 2 y −1 z − 5 = = 1 2 −4 y −1 z − 5 x−2 = = 2 −4
Persamaan Garis Simetrik
Latihan Soal 1. Cari suatu persamaan garis lurus melalui (3,2,1) dan
sejajar dengan vektor (3i-2j+6k)! 2. Cari persamaan garis lurus yang melalui titik (3,0-5) dan sejajar dengan garis r = (2,1,-5) + (0,-5,1)t !
Persamaan Bidang z N = ai+bj+ck
B(x,y,z) A(x0,y0,z0)
y
x
AB=(x AB=(x--x0)i+(y +(y--y0)j+(z +(z--z0)k N = ai ai + bj bj + ck ck Lakukan dot product antara AB dan N o NAB = NABcos 90 = 0 (a (aii+bj +bj+zk +zk)[(x )[(x--x0)i+(y +(y--y0)j+(z +(z--z0)k]=0 a(x a(x--x0)+b(y )+b(y--y0)+c(z )+c(z--z0)=0
ax+by+cz=ax0+by0+cz0
Yang diperlukan minimal: 1. 2.
Vektor normal bidang (N (N) Suatu titik pada bidang Jika diketahui 3 titik pada bidang bisa juga. Catatan: Jika suatu garis sejajar dengan arah bidangnya, maka θ=0.
N Garis
Catatan: Arah bidang selalu tegak lurus terhadap bidang
Contoh Soal: 1.
Tentukan persamaan bidang yang mencakup 3 titik A=(0,1,1); B=(2,1,3); C=(4,2,1) N
C=(4,2,1)
θ
A=(0,1,1)
AB=B-A AB=(2,1,3)-(0,1,1) AB=(2,0,2)
B=(2,1,3)
AC=C-A AC=(4,2,1)-(0,1,1) AC=(4,1,0)
N=ABxAC N=(2,0,2)x(4,1,0) N= i j k 2 4
0 1
2 0
N=0+8 =0+8jj+2k +2k+0+0-2i+0 N=-2i+8j+ +8j+2 2k a=-2, b=8, c=2
Lanjutan… Solusi Titik yang ditinjau A=(0,1,1) x0=0; y0=1; z0=1 ax+by+cz= ax0+by0+cz0 -2x+8y+2z=8+2 -2x+8y+2z=10
Latihan Soal: 1.
Cari persamaan bidang melalui titik (1,--1,0) dan sejajar dengan garis (1, r=(5 =(5i+j i+j--2k)+(2 )+(2ii-j+k)t !
