Jumat, 13 November 2009

Tugas Matematika Vektor Kelas XII-IPA

Kerjakan uji kompetensi Bab 4 Vektor hal. 214 no.12, 14, 15, 19, 20, 23, 27, 29.
Dikumpulkan pada hari jumat tgl. 20 Novemeber 2009.Tugas dikerjakan dalam blog ini...oke1
Diposting oleh di

37 komentar:

  1. Naritha16 November 2009 pukul 19.14

    Komentar ini telah dihapus oleh administrator blog.

    BalasHapus
  2. Anonim18 November 2009 pukul 19.07

    12. A(4, -1, -2), B(-6, 4,3), dan C(2,3,5). Titik P membagi AB sehingga AP : PB = 3 : 2 ?

    P = 2a + 3b / 2 + 3
    = 1/5 ( 2a + 3b )
    P = 1/5 .2 ( 4 ) 3(-6)
    (-1 ) + ( 4 )
    (-2 ) ( 3 )
    = 1/5 ( 8 ) (-18 )
    (-2) + ( 12 )
    (-4) ( 9 )

    = 1/5 (-10)
    ( 10)
    ( 5 )
    P = (-2)
    ( 2 )
    ( 1 )

    PC = C – P
    =( 2 ) (-2)
    ( 3 ) - ( 2 )
    ( 5 ) ( 1 )

    = ( 4 )
    ( 1 )
    ( 4 )


    14. a (x, 4, 7) b (6, y, 14). Nilai x-y=

    B = k. A

    (6) k.(x)
    (Y) = (4)
    (14) (7)


    14= k.7
    K = 14/7
    = 2
    6 = 2-x
    X = 3

    Y = 2.4
    = 8
    X – y = 3 – 8
    = -5

    15. a dan b sudut 60° |a|= 2 , |b|= 5, maka a(b+a)=

    = a(b+a) cosØ
    = 2 (5+2) cos 60°
    = 2 . 7 . ½
    = 14 . ½
    = 7

    19. a = 3xî + xð + k , b = -2î + 4ð + 5k , c = -3î + 2ð + k,Maka -3î + 2ð + k, maka a – c =
    A . b = (3x) (-2)
    ( x) . ( 4)
    (-4) ( 5)
    = 3x (-2) + x.4 + (-4).5
    = -6x + 4x + 20
    = -2x - 20
    -x = 10
    x = -10

    A - c = 3x + x - 4
    3.(-10)+ (-10)- 4
    = -30 -10 -4
    -3 + 2 +1
    = -33 - 12-5


    20. A( 2, -1, 4 ), B( 4, 1, 3 ), C(2, 0 ,5 ) kosinus sudut a dan b?

    Cos Ø = AB . AC/|AB|² . |AC|²

    AB = B - A
    (4) (2)
    (3) - (-1)
    (1) (4)
    = (2)
    (2)
    (-1)

    AC = C - A
    (2) (2)
    (0) - (-1)
    (5) (4)
    = (0)
    (1)
    (1)

    COS Ø = (2) (0)
    (2) .(1)
    (-1) (1)/AKAR 2²+2²+1 .AKAR 0²+1²+1²

    = 0+2-1 / AKAR 9 . AKAR 2
    = 1/3 AKAR 2

    27. panjang proyeksi a( -2, 8, 4), b( 0, p, 4) adalah 8, nilai p=

    AB = a . b
    = (-2) ( 0)
    ( 8) . ( P)
    ( 4) ( 4)/ AKAR 0²+P²+4²
    = 0 +8P+16/ AKAR P²+16

    |ab| = 8
    8(P+2)/ AKAR P²+16 = 8
    (P+2)²/ AKAR (P²+16)² = 1

    P²+4P+4/ P²+16 = 1

    (P²+4P+4)= (P²+16)
    4P - 12
    4(P-3)
    P=3

    AZHAR FITRIYATI
    XII IPA

    BalasHapus
  3. Anonim18 November 2009 pukul 19.11

    12. A(4, -1, -2), B(-6, 4,3), dan C(2,3,5). Titik P membagi AB sehingga AP : PB = 3 : 2 ?

    P = 2a + 3b / 2 + 3
    = 1/5 ( 2a + 3b )
    P = 1/5 .2 ( 4 ) 3(-6)
    (-1 ) + ( 4 )
    (-2 ) ( 3 )
    = 1/5 ( 8 ) (-18 )
    (-2) + ( 12 )
    (-4) ( 9 )

    = 1/5 (-10)
    ( 10)
    ( 5 )
    P = (-2)
    ( 2 )
    ( 1 )

    PC = C – P
    =( 2 ) (-2)
    ( 3 ) - ( 2 )
    ( 5 ) ( 1 )

    = ( 4 )
    ( 1 )
    ( 4 )
    14. a (x, 4, 7) b (6, y, 14). Nilai x-y=

    B = k. A

    (6) k.(x)
    (Y) = (4)
    (14) (7)


    14= k.7
    K = 14/7
    = 2
    6 = 2-x
    X = 3

    Y = 2.4
    = 8
    X – y = 3 – 8
    = -5

    15. a dan b sudut 60° |a|= 2 , |b|= 5, maka a(b+a)=

    = a(b+a) cosØ
    = 2 (5+2) cos 60°
    = 2 . 7 . ½
    = 14 . ½
    = 7

    19. a = 3xî + xð + k , b = -2î + 4ð + 5k , c = -3î + 2ð + k,Maka -3î + 2ð + k, maka a – c =
    A . b = (3x) (-2)
    ( x) . ( 4)
    (-4) ( 5)
    = 3x (-2) + x.4 + (-4).5
    = -6x + 4x + 20
    = -2x - 20
    -x = 10
    x = -10


    Rahmi Indah P
    XII IPA

    yg lain na blum bu......
    soal na simbol na gg bsa, di komputer saya gg ada simbol yg buat kerjain matematika......

    A - c = 3x + x - 4
    3.(-10)+ (-10)- 4
    = -30 -10 -4
    -3 + 2 +1
    = -33 - 12-5

    BalasHapus
  4. Naritha_XII IPA19 November 2009 pukul 16.37

    bu saya udah coba pake Microsoft Equation 3.0.
    tapi gak bisa ke baca di komentar!!
    soalnya Microsoft Equation 3.0 itu bukan dalam bentuk text.
    gimana dong?
    mending lewat e-mail aja ya bu nanti kalo udah di kirim lewat e-mail, nanti saya ketik di komentar kalo saya udah kirim lewat e-mail..atau gak di print out?

    Naritha

    BalasHapus
  5. Siti Farida Faradina19 November 2009 pukul 16.38

    Nama : Siti Farida Faradina
    Kelas : XII-IPA


    12) = 3 : 2

    P = 2a+3b = 1/5
    = 1/5 . {2.(4,-1,-2) + 3.(-6,4,8)
    = 1/5 (8,-2,-4) + (-18,12,9)
    P = 1/5 (-10,10,5) ℮

    14) B = K.A B = K.x
    (6,y,14) = K (x,4,7) 6 = 2.x
    14 = K.7 x = 6/2
    K = 14/7 x = 3
    = 2
    Y = K.4
    = 2.4
    = 8
    x-y = 3-8
    = -5 a

    15) a.(b+a) cos θ
    = 2.(5+2) cos 60°
    = 7 b


    20) cos θ =
    = B-A
    = (4,1,3) – (2,-1,4)
    = (2,2,-1)
    = C-A
    = (2,1,5) – (2,-1,4)
    = (0,1,1)

    Cos θ =
    =
    = d

    23) cos =
    |OA| |AB|
    = 0 + 1 + 2
    .
    = .

    . =
    =
    =
    = 3 c
    27)
    =

    =
    =


    Dik ; sehingga dapat dibentuk persamaan
    = 8
    = 1


    4+p+p² = 16+p²
    4p-12
    4(p-3)
    P = 3 c

    29) =
    =

    =

    =
    =
    = 4 i + 6j – 2k d

    BalasHapus
  6. ADI PUTRA19 November 2009 pukul 17.39

    bu gmna nihh....
    saya bingung...

    saya sudah mengerjakan di msWord,
    tapi gak bisa di copy di blogger ibu

    sedangkan di blogger ibu gak da fasilitasnya

    gmn donkkk.......?????

    help me......

    BalasHapus
  7. Anonim19 November 2009 pukul 17.56

    No 12

    → →

    AР : PB =3:2

    P=2a+3b 1

    = (2a+3b)

    2+3 5

    1

    Р= ((4﴿ )6)

    (2(-1)+3 (4)

    ((-2) (3)



    1 ( 8) (-18)

    =  (-12)+ ( 12)

    5 (-4) ( 9)

    (e)

    (-2)

    Р=( 2)

    ( 1)



    → → →

    Рc c  Р

    2 -1 (4﴿

    =(3}  {2}=(1﴿

    5 1 (4﴿

    no14



    B =K.A

    (6 ) (X)

    (Y )=K (4) ﴾A﴿

    (14) (7)



    14=K.7 6=2X B =K.A X-Y

    2 =K 3= X Y=2.4 3-8

    Y=8 -5





    NO 15

    a (b+a)cos60°

    2 (5+2)1/2 ﴾B﴿

    2.7.1/2

    14.1/2

    7





    NO 19



    A B

    ﴾ 3X﴿ (2-)

    ﴾ X﴿ x (4 ) ﴾=­2﴿.3X+X.4+﴾­4﴿.5

    (-4) ( 5) ═-6X+4X-20

    ═ -2X-20

    -20═2X

    -10═X



    A═3X+X-4 A-C═-30Ǐ-10ĵ-4ќ-﴾-3Ǐ+2ĵ+ќ﴿ ﴾E﴿

    ═ 3﴾-10﴿+﴾-10﴿-4 ═-33Ǐ-12ĵ-5ќ

    ═-30-10-4



    No20



    Cosθ═ → →



    AB AC

     . 

    2 2

    |AB| |AC|

    → →

    AB ═B . A AC=C - A

    (4) ( 2) ( 2) (2) ( 2) (0)

    (1) - (-1) = ( 2) (0) - (-1) = (1)

    (3) ( 4) (-1) (5) ( 4) (1)



    ( 2) (0)

    COS θ=( 2) . (1)

    (-1) (1)

    

    √2²+2²+1² .√0²+1²+1² (D)



    0+2-1 1

    ═ ═



    √9√2 3√2





    NO 27

    a.b

    AB=

    |b|







    =(-2) (0)

    (8 ) . (P)

    ( 4) (4) 0+8P+16

     = = 4(A)

    √0²+P²+4 √P²+16

    BalasHapus
  8. Irfan Setiaone19 November 2009 pukul 18.02

    bu...,, gak bisa di paste dari ms word...
    saya udh ngrjain d' Ms word..
    gmn bu..??

    BalasHapus
  9. Naritha_XII IPA19 November 2009 pukul 18.03

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)

    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3

    B=K.A
    Y=2.4
    Y=8

    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10

    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12

    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d)

    BalasHapus
  10. Naritha_XII IPA19 November 2009 pukul 18.14

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)

    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3

    B=K.A
    Y=2.4
    Y=8

    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10

    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12

    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d)

    BalasHapus
  11. Anonim19 November 2009 pukul 19.00

    Anggri mutiara cahya
    XII-ipa


    12. A = (4 , -1 , -2)
    B = (-6 , 4 , 3)
    C = (2 , 3 5)

    → →
    AP : PB = 3 : 2
    P = 2A + 3B
    ————――
    2 + 3
    = 1/5 (2a+3b)

    P =1/5.2(4,-1,2)+3(-6,4,3)
    =1/5(8,-2,-4)+(-18,12,9)
    =1/5 (-10 , 10 , 5)
    =(-2,2,1)
    →
    PC =C-P
    =(2,3,5)-(-2,2,1)
    =(4,1,4)


    14. B = K A
    (6,Y,14)=K(X,4,7)
    14 = K.7
    K = 14/7
    K = 2 6 = 2 - X
    Y = 2.4 X = 3
    Y = 8
    X-Y = 3-8 = -5


    15. a (b+a) cos 60
    = 2(5+2) ½
    = 2(5) + 2(2) ½
    = 10+4 ½
    = 14 ½
    = 7

    →
    19. AB = b – a
    (3X , X , -4) (-2 , 4 , 5)
    = 3x – (2) + x . 4 + (-4) 5
    = -6x + 4x – 20
    = -2x – 20
    20 = -2x
    20/-2 = x
    - 10 = x

    [3x+4-4]
    [3-10+(10)-4]
    [(-30)-10-4]
    [(-3)+2-1]
    ————————
    = -33-12-5


    → →
    20. cos Θ = AB . AC
    —―—―—―—
    [AB].[AC]
    →
    AB = b – a
    = (4 , 1 , 3) – (2 , -1 , 4)
    = (2 , 2, -1)
    →
    AC = c – a
    = (2 , 0 , 5) – (2 , -1 , 4)
    = (0 ,1 , 1)

    cos Θ = (2 , 2 , -1) . (0 , 0 , 1)
    = √(2²+2²+1).(0²+1²+1²)
    = 0+2-1
    —―—―
    √9.√2
    = 1
    ――
    3√2


    23. Cos θ OA.AB = 0+1+2
    ―—―—― —―—―
    |OA||AB| √6.√2
    = 3
    ―—
    √12
    → →
    OA.AB =|OA||AP|Cos θ
    =√1+1+4.√1+4+9.2
    —―
    √12
    =√6.√14.3
    ―—
    √12
    = 3√7


    27. |Ab| = a.b=(-2,8,4).(0,P,4)
    —―— ―—―――—――—―—
    |B| √(0)²+(p)²+(4)²
    = (-2).0+8p+4.4
    ―—―—―—―—
    √16+p²
    = (16+8p)
    ―—―—―
    √16+p²

    |Ab| = 8
    8(2+p)= 8
    —―—―—
    √16+p²
    (2+p)²= 1
    ―—―—―
    √16+p²
    4+4p+p²
    —―—―— = 1
    16+p²
    4+4p+p²= 16+p²
    = 4p-12
    = 4(p-3)
    p = +3




    29. Uv = U.v .V
    ―—―
    |V|²
    = 28 (2,3,-1)
    ―—―—―—―—―—
    √((2)²+(3)²+(-1)²)²
    = 28/14 (2 , 3 , -1)
    = 56/14 î + 84/14 ĵ + -28/4 K
    = 8/2 î + 12/2 ĵ + -4/2 K
    = 4 ĵ + 6 î – 2 K

    BalasHapus
  12. Anonim19 November 2009 pukul 19.01

    marissa rahmadiningtyas
    XII-ipa



    12. A = (4 , -1 , -2)
    B = (-6 , 4 , 3)
    C = (2 , 3 5)

    → →
    AP : PB = 3 : 2
    P = 2A + 3B
    ————――
    2 + 3
    = 1/5 (2a+3b)

    P =1/5.2(4,-1,2)+3(-6,4,3)
    =1/5(8,-2,-4)+(-18,12,9)
    =1/5 (-10 , 10 , 5)
    =(-2,2,1)
    →
    PC =C-P
    =(2,3,5)-(-2,2,1)
    =(4,1,4)


    14. B = K A
    (6,Y,14)=K(X,4,7)
    14 = K.7
    K = 14/7
    K = 2 6 = 2 - X
    Y = 2.4 X = 3
    Y = 8
    X-Y = 3-8 = -5


    15. a (b+a) cos 60
    = 2(5+2) ½
    = 2(5) + 2(2) ½
    = 10+4 ½
    = 14 ½
    = 7

    →
    19. AB = b – a
    (3X , X , -4) (-2 , 4 , 5)
    = 3x – (2) + x . 4 + (-4) 5
    = -6x + 4x – 20
    = -2x – 20
    20 = -2x
    20/-2 = x
    - 10 = x

    [3x+4-4]
    [3-10+(10)-4]
    [(-30)-10-4]
    [(-3)+2-1]
    ————————
    = -33-12-5





    → →
    20. cos Θ = AB . AC
    —―—―—―—
    [AB].[AC]
    →
    AB = b – a
    = (4 , 1 , 3) – (2 , -1 , 4)
    = (2 , 2, -1)
    →
    AC = c – a
    = (2 , 0 , 5) – (2 , -1 , 4)
    = (0 ,1 , 1)

    cos Θ = (2 , 2 , -1) . (0 , 0 , 1)
    = √(2²+2²+1).(0²+1²+1²)
    = 0+2-1
    —―—―
    √9.√2
    = 1
    ――
    3√2


    23. Cos θ OA.AB = 0+1+2
    ―—―—― —―—―
    |OA||AB| √6.√2
    = 3
    ―—
    √12
    → →
    OA.AB =|OA||AP|Cos θ
    =√1+1+4.√1+4+9.2
    —―
    √12
    =√6.√14.3
    ―—
    √12
    = 3√7


    27. |Ab| = a.b=(-2,8,4).(0,P,4)
    —―— ―—―――—――—―—
    |B| √(0)²+(p)²+(4)²
    = (-2).0+8p+4.4
    ―—―—―—―—
    √16+p²
    = (16+8p)
    ―—―—―
    √16+p²

    |Ab| = 8
    8(2+p)= 8
    —―—―—
    √16+p²
    (2+p)²= 1
    ―—―—―
    √16+p²
    4+4p+p²
    —―—―— = 1
    16+p²
    4+4p+p²= 16+p²
    = 4p-12
    = 4(p-3)
    p = +3




    29. Uv = U.v .V
    ―—―
    |V|²
    = 28 (2,3,-1)
    ―—―—―—―—―—
    √((2)²+(3)²+(-1)²)²
    = 28/14 (2 , 3 , -1)
    = 56/14 î + 84/14 ĵ + -28/4 K
    = 8/2 î + 12/2 ĵ + -4/2 K
    = 4 ĵ + 6 î – 2 K

    BalasHapus
  13. Anonim19 November 2009 pukul 19.03

    panca ratna koli
    XII-ipa




    12. A = (4 , -1 , -2)
    B = (-6 , 4 , 3)
    C = (2 , 3 5)

    → →
    AP : PB = 3 : 2
    P = 2A + 3B
    ————――
    2 + 3
    = 1/5 (2a+3b)

    P =1/5.2(4,-1,2)+3(-6,4,3)
    =1/5(8,-2,-4)+(-18,12,9)
    =1/5 (-10 , 10 , 5)
    =(-2,2,1)
    →
    PC =C-P
    =(2,3,5)-(-2,2,1)
    =(4,1,4)


    14. B = K A
    (6,Y,14)=K(X,4,7)
    14 = K.7
    K = 14/7
    K = 2 6 = 2 - X
    Y = 2.4 X = 3
    Y = 8
    X-Y = 3-8 = -5


    15. a (b+a) cos 60
    = 2(5+2) ½
    = 2(5) + 2(2) ½
    = 10+4 ½
    = 14 ½
    = 7

    →
    19. AB = b – a
    (3X , X , -4) (-2 , 4 , 5)
    = 3x – (2) + x . 4 + (-4) 5
    = -6x + 4x – 20
    = -2x – 20
    20 = -2x
    20/-2 = x
    - 10 = x

    [3x+4-4]
    [3-10+(10)-4]
    [(-30)-10-4]
    [(-3)+2-1]
    ————————
    = -33-12-5





    → →
    20. cos Θ = AB . AC
    —―—―—―—
    [AB].[AC]
    →
    AB = b – a
    = (4 , 1 , 3) – (2 , -1 , 4)
    = (2 , 2, -1)
    →
    AC = c – a
    = (2 , 0 , 5) – (2 , -1 , 4)
    = (0 ,1 , 1)

    cos Θ = (2 , 2 , -1) . (0 , 0 , 1)
    = √(2²+2²+1).(0²+1²+1²)
    = 0+2-1
    —―—―
    √9.√2
    = 1
    ――
    3√2


    23. Cos θ OA.AB = 0+1+2
    ―—―—― —―—―
    |OA||AB| √6.√2
    = 3
    ―—
    √12
    → →
    OA.AB =|OA||AP|Cos θ
    =√1+1+4.√1+4+9.2
    —―
    √12
    =√6.√14.3
    ―—
    √12
    = 3√7


    27. |Ab| = a.b=(-2,8,4).(0,P,4)
    —―— ―—―――—――—―—
    |B| √(0)²+(p)²+(4)²
    = (-2).0+8p+4.4
    ―—―—―—―—
    √16+p²
    = (16+8p)
    ―—―—―
    √16+p²

    |Ab| = 8
    8(2+p)= 8
    —―—―—
    √16+p²
    (2+p)²= 1
    ―—―—―
    √16+p²
    4+4p+p²
    —―—―— = 1
    16+p²
    4+4p+p²= 16+p²
    = 4p-12
    = 4(p-3)
    p = +3




    29. Uv = U.v .V
    ―—―
    |V|²
    = 28 (2,3,-1)
    ―—―—―—―—―—
    √((2)²+(3)²+(-1)²)²
    = 28/14 (2 , 3 , -1)
    = 56/14 î + 84/14 ĵ + -28/4 K
    = 8/2 î + 12/2 ĵ + -4/2 K
    = 4 ĵ + 6 î – 2 K

    BalasHapus
  14. Anonim19 November 2009 pukul 19.03

    octaviani ayuningtyas
    XII-IPA



    12. A = (4 , -1 , -2)
    B = (-6 , 4 , 3)
    C = (2 , 3 5)

    → →
    AP : PB = 3 : 2
    P = 2A + 3B
    ————――
    2 + 3
    = 1/5 (2a+3b)

    P =1/5.2(4,-1,2)+3(-6,4,3)
    =1/5(8,-2,-4)+(-18,12,9)
    =1/5 (-10 , 10 , 5)
    =(-2,2,1)
    →
    PC =C-P
    =(2,3,5)-(-2,2,1)
    =(4,1,4)


    14. B = K A
    (6,Y,14)=K(X,4,7)
    14 = K.7
    K = 14/7
    K = 2 6 = 2 - X
    Y = 2.4 X = 3
    Y = 8
    X-Y = 3-8 = -5


    15. a (b+a) cos 60
    = 2(5+2) ½
    = 2(5) + 2(2) ½
    = 10+4 ½
    = 14 ½
    = 7

    →
    19. AB = b – a
    (3X , X , -4) (-2 , 4 , 5)
    = 3x – (2) + x . 4 + (-4) 5
    = -6x + 4x – 20
    = -2x – 20
    20 = -2x
    20/-2 = x
    - 10 = x

    [3x+4-4]
    [3-10+(10)-4]
    [(-30)-10-4]
    [(-3)+2-1]
    ————————
    = -33-12-5





    → →
    20. cos Θ = AB . AC
    —―—―—―—
    [AB].[AC]
    →
    AB = b – a
    = (4 , 1 , 3) – (2 , -1 , 4)
    = (2 , 2, -1)
    →
    AC = c – a
    = (2 , 0 , 5) – (2 , -1 , 4)
    = (0 ,1 , 1)

    cos Θ = (2 , 2 , -1) . (0 , 0 , 1)
    = √(2²+2²+1).(0²+1²+1²)
    = 0+2-1
    —―—―
    √9.√2
    = 1
    ――
    3√2


    23. Cos θ OA.AB = 0+1+2
    ―—―—― —―—―
    |OA||AB| √6.√2
    = 3
    ―—
    √12
    → →
    OA.AB =|OA||AP|Cos θ
    =√1+1+4.√1+4+9.2
    —―
    √12
    =√6.√14.3
    ―—
    √12
    = 3√7


    27. |Ab| = a.b=(-2,8,4).(0,P,4)
    —―— ―—―――—――—―—
    |B| √(0)²+(p)²+(4)²
    = (-2).0+8p+4.4
    ―—―—―—―—
    √16+p²
    = (16+8p)
    ―—―—―
    √16+p²

    |Ab| = 8
    8(2+p)= 8
    —―—―—
    √16+p²
    (2+p)²= 1
    ―—―—―
    √16+p²
    4+4p+p²
    —―—―— = 1
    16+p²
    4+4p+p²= 16+p²
    = 4p-12
    = 4(p-3)
    p = +3




    29. Uv = U.v .V
    ―—―
    |V|²
    = 28 (2,3,-1)
    ―—―—―—―—―—
    √((2)²+(3)²+(-1)²)²
    = 28/14 (2 , 3 , -1)
    = 56/14 î + 84/14 ĵ + -28/4 K
    = 8/2 î + 12/2 ĵ + -4/2 K
    = 4 ĵ + 6 î – 2 K

    BalasHapus
  15. Irfan Setiawan XII IPA19 November 2009 pukul 19.08

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))

    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4

    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7

    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  16. adi putra xii ipa19 November 2009 pukul 19.12

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))
    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4
    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7
    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  17. aa19 November 2009 pukul 19.18

    ....

    BalasHapus
  18. Rinanda Nur Rizki XII - IPA19 November 2009 pukul 19.21

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))
    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4
    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7
    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  19. Rio bagus19 November 2009 pukul 19.29

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))
    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4
    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7
    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  20. Rama OB XII IPA19 November 2009 pukul 19.35

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))
    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4
    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7
    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  21. Richo HS19 November 2009 pukul 19.35

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))
    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4
    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7
    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  22. M. AJIL19 November 2009 pukul 19.37

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))
    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4
    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7
    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  23. Angri Syaiful F19 November 2009 pukul 19.37

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))
    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4
    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7
    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  24. .19 November 2009 pukul 19.44

    .

    BalasHapus
  25. Sinta Setianti19 November 2009 pukul 19.45

    12. AP : BP . 3:2
    P=(2a+3b)/(2+3)=1/5.( 2a+3b)
    A=■(4@-1@-2) b=■(-6@4@3)
    p= 1/5 [2 ■(4@-1@-2)+3■(-6@4@3)]
    = 1/5 [■(8@-2@-4)+■(-18@12@9)]
    = 1/5 (■(-10@10@5))+(■(-2@2@1))
    PC = C-P
    =(■(2@3@5))-(■(-2@2@1))
    =(■(4@1@4))
    14. AB = B-A
    =(■(6@y@14))-(■(-x@a@7))
    =(■(6-x@y-4@7))
    6 -x= 0
    -x= -6
    X= -6
    x-9=-6.(4)
    = -2
    y-4=0
    y=4
    15. a.(b+a)cosѳ
    2(5+2)cos 60
    2(7).1/2
    = 14 1/2
    = 7
    19. AB = B-A
    =(■(3x@x@-4)).(■(-2@4@5))
    = 3x(-2)+x.4+(-4).5
    = -6x+4x-20
    -2x=20
    -2x=120
    -x=10
    X=-10
    3x+x-4
    3-10+(-10)(-4)
    -30-10-4
    (-3+2-1)/(-33-12-5)+

    20. cosѳ = (AB AC)/█(|AB|2 |AC|2@)
    AB = B-A
    =(■(4@1@3))-(■(2@-1@4))=(■(2@2@1))
    AC = C-A
    =(■(2@0@5))-(■(2@-1@4))= (■(0@1@1))
    Cosѳ= (■(2@2@-1))(■(0@1@1))
    =√(2kuadrat+2kuadrat+1.) √0kudrat+1kudrat+1kudrat
    =(0+2-1)/(√9 √2)
    = 1/3 √2

    BalasHapus
  26. Anonim19 November 2009 pukul 19.54

    No 12
    → →
    AР : PB =3:2
    P=2a+3b 1
    = (2a+3b)
    2+3 5
    1
    Р= ((4﴿ )6)
    (2(-1)+3 (4)
    ((-2) (3)

    1 ( 8) (-18)
    =  (-12)+ ( 12)
    5 (-4) ( 9)
    (e)
    (-2)
    Р=( 2)
    ( 1)

    → → →
    Рc c  Р
    2 -1 (4﴿
    =(3}  {2}=(1﴿
    5 1 (4﴿
    no14

    B =K.A
    (6 ) (X)
    (Y )=K (4) ﴾A﴿
    (14) (7)

    14=K.7 6=2X B =K.A X-Y
    2 =K 3= X Y=2.4 3-8
    Y=8 -5


    NO 15
    a (b+a)cos60°
    2 (5+2)1/2 ﴾B﴿
    2.7.1/2
    14.1/2
    7


    NO 19

    A B
    ﴾ 3X﴿ (2-)
    ﴾ X﴿ x (4 ) ﴾=­2﴿.3X+X.4+﴾­4﴿.5
    (-4) ( 5) ═-6X+4X-20
    ═ -2X-20
    -20═2X
    -10═X

    A═3X+X-4 A-C═-30Ǐ-10ĵ-4ќ-﴾-3Ǐ+2ĵ+ќ﴿ ﴾E﴿
    ═ 3﴾-10﴿+﴾-10﴿-4 ═-33Ǐ-12ĵ-5ќ
    ═-30-10-4

    No20

    Cosθ═ → →

    AB AC
     . 
    2 2
    |AB| |AC|
    → →
    AB ═B . A AC=C - A
    (4) ( 2) ( 2) (2) ( 2) (0)
    (1) - (-1) = ( 2) (0) - (-1) = (1)
    (3) ( 4) (-1) (5) ( 4) (1)

    ( 2) (0)
    COS θ=( 2) . (1)
    (-1) (1)
    
    √2²+2²+1² .√0²+1²+1² (D)

    0+2-1 1
    ═ ═

    √9√2 3√2


    NO 27
    a.b
    AB=
    |b|



    =(-2) (0)
    (8 ) . (P)
    ( 4) (4) 0+8P+16
     = = 4(A)
    √0²+P²+4 √P²+16

    No 23

    Cos θ= 0A.AB 3 1 0
     =  (1) (1)
    √6. √2 √12 2 1


    → →
    OA . AP=|OA| . |AP| Cos θ
    √1+1+4 √1+4+9 3
    
    √12
    =√6.√14. 3
    
    √12
    =3√7


    ivone.angelia
    XI-IPA

    BalasHapus
  27. Mia Alyunas19 November 2009 pukul 20.45

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)

    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3

    B=K.A
    Y=2.4
    Y=8

    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10

    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12

    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d).

    BalasHapus
  28. Sri Bintang Wahyuni19 November 2009 pukul 20.46

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)

    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3

    B=K.A
    Y=2.4
    Y=8

    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10

    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12

    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d).

    BalasHapus
  29. Yudha Afrizal_XII IPA19 November 2009 pukul 20.46

    12.
    AP:Pb=3:2
    P=2a+3b/2+3=1/5(2a+3B)
    =1/5(2(4,-1,-2)+3(-6,4,3)
    =1/5(8,-2,-4)+(-18,12,9)
    =1/5(-10,10,5)
    =(-2,2,1)
    PC=C-P
    =(2,3,5)-(-2,2,1)
    =(4,1,4)

    14.
    a=k.b
    (x,4,7)=1/2(6,y,14)
    x=1/2.6
    x=3

    4=1/2y
    y=8

    =x-y
    =3-8
    =-5

    27.
    (-2,8,4) (0,P,4)
    (akar)0kuadrat+Pkuadrat+4kuadrat
    (akar)16+Pkuadrat
    4+P=8
    P=8-4
    P=4

    BalasHapus
  30. Videa XII IPA19 November 2009 pukul 20.47

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)

    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3

    B=K.A
    Y=2.4
    Y=8

    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10

    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12

    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d).

    BalasHapus
  31. Fitria Permata Sari19 November 2009 pukul 20.49

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)

    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3

    B=K.A
    Y=2.4
    Y=8

    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10

    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12

    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d).

    BalasHapus
  32. mia al-yunas19 November 2009 pukul 20.50

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)
    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3
    B=K.A
    Y=2.4
    Y=8
    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10
    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12
    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d).

    BalasHapus
  33. Mia Alyunas19 November 2009 pukul 20.50

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)

    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3

    B=K.A
    Y=2.4
    Y=8

    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10

    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12

    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d).

    BalasHapus
  34. widya noviela dewi19 November 2009 pukul 20.51

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)
    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)
    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3
    B=K.A
    Y=2.4
    Y=8
    X-Y=3-8= -5 (a)
    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)
    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10
    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)
    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)
    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12
    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)
    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)
    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d).

    BalasHapus
  35. Welly Nugraha_XII IPA19 November 2009 pukul 20.58

    12.
    AP:PB=3:2
    P=2a+3b/2+3=1/5(2a+3b)
    =1/5(2(4,-1,-2)+3(-6,4,3)
    =1/5(8,-2,-4)+(-18,12,9)
    =1/5(-10,10,5)
    =(-2,2,1)
    PC=C-P
    =(2,3,5)-(-2,2,1)
    =(4,1,4)

    14.
    a=k.b
    (X,4,7)=1/2(6,y,14)
    X=1/2.6
    X=3

    4=1/2y
    y=8

    =X-y
    =3-8
    =-5

    27.
    (-2,8,4) (0,P,4)
    (akar)0kuadrat+Pkuadrat+4kuadrat
    (akar)16+Pkuadrat
    4+P=8
    P=8-4
    P=4

    BalasHapus
  36. Barka Rosyad_XII IPA19 November 2009 pukul 21.10

    12.
    AP:PB=3 : 2
    P=2a+3b/2+3=1/5(2a+3b)
    =1/5(2(4,-1,-2)+3(-6,4,3)
    =1/5(8,-2,-4)+(-18,12,9)
    =1/5(-10,10,5)
    =(-2,2,1)
    PC=C – P
    =(2,3,5)-(-2,2,1)
    =(4,1,4)
    No.14)
    a=k.b
    (x,4,7)=1/2(6,y,14)
    x=1/2.6
    x=3

    4=1/2y
    y=8

    =x-y
    =3-8
    =-5]

    27.
    (-2,8,4)(0,P,4)
    (akar)0kuadrat+Pkuadrat+4kuadrat
    (akar)16+Pkuadrat
    4+P=8
    P=8-4
    P=4

    BalasHapus
  37. Sartika Burhan20 November 2009 pukul 05.33

    12) AP:PB = 3:2
    P=2a+3b/2+3 = 1/5 (2a+3b)
    P=1/5 (2(4,-1,-2)) + (3(-6,4,3))
    P=1/5 (8,-2,-4) + (-18,12,9)
    P=1/5(-10,10,5)
    P=(-2,2,1)

    PC=C-P
    PC=(2,3,5)-(-2,2,1)=(4,1,4) (e)

    14) B=K.A
    (6,-1,14)=K.(x,4,7)
    14=K.7
    2=K
    6=2x
    X=3

    B=K.A
    Y=2.4
    Y=8

    X-Y=3-8= -5 (a)

    15) a(b+a) cos 60o
    =2(5+2).1/2
    =2.7.1/2
    =14.1/2= 7 (b)

    19) A.B=(3x,x,-4)(-2,4,5)
    =(3x.-2)+(x.4)+(-4.5)
    = -6x+4x-20
    = -2x-20
    = x=-10

    A-C=(3(-10),-10,-4)-(-3,2,1)
    =(3(-10)+3,-10-2,-4-1)
    =(-27,-12,-5)
    = -27i -12j -5k (d)

    20) cos ‘teta’= AB.AC/|AB|2.|AC|2
    AB=B-A
    = (4,1,3)-(2,-1,4)=(2,2,-1)
    AC=C-A
    =(2,0,5)-(2,-1,4)=(0,1,1)
    Cos ‘teta’= (2.0)+(2.1)+(-1.1)/(akar 22.22.(-1) 2). (akar 0.12.1 2)
    = 2-1/(akar 9).(akar 2)
    1/3akar2 (d)

    23) cos ‘teta’= OA.AB/|OA||AB|
    =0+1+2/(akar 6).(akar 2)
    =3/akar 12

    OA.AP=|OA||AP| cos ‘teta’
    = (akar 1+1+4) (akar 1+4+9). 3/akar 12
    = (akar 6).(akar 14).3/akar 12
    = 3 akar 7 (b)

    27) ab=a.b/|b|
    =(-2.0)+(8.P)+(4.4)/akar 0+2+4
    =8P+16/akar P2+16
    |ab|=8
    8(P+2)/akar P2+16=8
    =(P+2)2/(akar P2+16)2=1
    =P2+4P+4/P2+16=1
    = P2+4P+4=P2+16
    4P-12=0
    4P=12
    P=3 (c)

    29) Va=U.V/|V|2.V
    =(6.2)+(3.3)+(-7.-1)/(akar 22+32+ (-1)2)2.(2,3,-1)
    =12+9+7/(akar) 2.(2,3,-1)
    =28/14 . (2,3,-1)
    =2. (2,3,-1)
    =(4,6,-2)
    =4i+6j-2k (d)

    BalasHapus

Langganan: Posting Komentar (Atom)