PTS

JAWABAN: 

1. vektor  a+2b-3c

     = (  1 2 3 ) + 2 ( 5 4 -1 ) - 3 ( 4 -1 1) 

     = ( 1 2 3 ) + (10 8 -2) - ( 12 -3  3 ) 

     = [1 2 3 + 10 8 (-2) - 12 (-3) 3]

     = [-1 13 -2]

= (-1   13   -2 ) [D] 

2. |a|= akar 3 

    |b|= 1

    |a-b| = 1 

|a-b|^2 = |a|^2 + |b|^2 - 2.a.b cos a 

1^2= ( akar 3 )^2 + 1^2 ab cos a

1= 3+1-2 ab cos a 

1= 4-2ab cos a 

2ab cos a = 4-1 

2ab cos a = 3 

---------------

|a+b|^2 = |a|^2 + |b|^2 + 2ab cos a 

|a+b|^2= ( akar 3)^2 + 1^2 + 3

|a+b|^2= 3+1+3 

|a+b|^2 = 7 

|a+b| = akar 7 [C]

3. a= 2i - 3j + 4k

    b= 5j+5k= 0i+5j+5k

   ab= 2(0) + (-3)(5) + 4(5) = 0-15+20= 5  [akar 5] [B]

4. |a + b| = 2√19

|a + b|^2 = (2√19)^2
|a|^2 + 2ab + |b|^2 = 4(19)
4^2 + 2ab + 6^2 = 76
16 + 2ab + 36 = 76
2ab = 24

|a - b|^2 = |a|^2 - 2ab + |b|^2
|a - b|^2 = 4^2 - 24 + 6^2
|a - b|^2 = 28
|a - b| = √28 = 2√7 [A] 

5. b.c=0 

(p,2,-1) . (1,-1,3)=0

p-2-3=0

p-5=0

p=5 

jadi b= 5i+2j-k

vektor a-b-c

= (2,-3,1) - (5,2,-1) - (1,-1,3)

= ( 2-5-1, -3-2-(-1), 1-(-1)-3)

= (-4,-4,-1)

= -4i-4j-k [C]

6. a . b = |a| |b| Cos 60
p² - 1 = √(p² + 3) √(p² + 3) (1/2)
2p² - 2 = p² + 3
p² = 5
P² - 5 = 0
(P + √5)(p - √5) = 0
P = √5
P = - √5 [D] 

7. mencari vektor AB dan BC

AB= (1, -2,1) - (3,2,-1) = (-2, -4, 2)

BC= (7p, -1, -5) - ( 1, -2, 1) = ( 6p + 1 -6)

mencari nilai p 

BC=k.AB 

[6 p+1 -5) =k (-2 -4 2) 

6=-2.k

k=-3 

maka 

p+1=k.(-4)

p+1= (-3).(-4)

p+1= 12 ~> p=11 [D]

NO 8. 

9. 

a . b = 

a . b = –2(0) + 8p + 4(4)

a . b = 8p + 16

Panjang vektor b

|b| = 

|b| = 

Panjang proyeksi vektor a pada b = 8

 = 8

 = 8

 = 8

8(p + 2) = 8√(p² + 16)

(p + 2) = √(p² + 16)

==> kedua ruas dikuadratkan <==

(p + 2)² = (p² + 16)

p² + 4p + 4 = p² + 16

4p = 16 – 4

4p = 12

p = 3 [C]

10. a . b = 0

(p, 2, -1) . (4, -3, 6) = 0
p(4) + 2(-3) + (-1)(6) = 0
4p - 6 - 6 = 0
4p = 12
p = 3

a - 2b
= (p, 2, -1) - 2(4, -3, 6)
= (3, 2, -1) - (8, -6, 12)
= (-5, 8, -13)

3c = 3(2, -1, 3) = (6, -3, 9)

(a - 2b) . 3c
= (-5, 8, -13) . (6, -3, 9)
= -5(6) + 8(-3) + (-13)(9)
= -30 - 24 - 117
= -171 [E]

11. AB = b - a
     = (3,3,1) - (1,2,3)
     = (2,1,-2)

BC = c - b
     = (7,5,-3) - (3,3,1)
     = (4,2,-4)

      AB : BC
(2,1,-2) : (4,2,-4)
(2,1,-2) : 2(2,1,-2)
       1  :  2

Jadi, AB : BC = 1 : 2 [A]

12. |a+b| = √|a|² + |b|² + 2 |a| |b| . cos x

|a-b| = √|a|² + |b|² - 2 |a| |b| . cos x

√|a|² + |b|² + 2 |a| |b| . cos x = √|a|²+|b|² + 2 |a| |b| . cosx

AKAR NYA HILANG JADI

|a|² + |b|² + 2 |a| |b| . cosx = |a|²+|b|² + 2 |a| |b| . cosx

=> |a| |b| . cos x + 2 |a| |b| . cos x = 0

=> 4 |a| |b| . cos x = 0

=> 0/4 |a|.|b|

=> 0

cos x = 0 = 90°

jadi a dan b saling membentuk sudut 90° [A]

13. AB= b-a

BC= c-b

AB=(-1,1,-1)-(2,7,8)
=(-3,-6,-9)

BC=(0,3,2)-(-1,1,-1)
=(1,2,3)

e= u•v/|v|² kali v
=(-3•1)+(-6•2)+(-9•3)/(√1²+2²+3²)²kali (1,2,3)
=-3-12-27/(√14)² kali (1,2,3)
=-42/14 kali (1,2,3)
= -3(1,2,3)
= (-3i,-6j,-9k) [A]

14.  a . b = 0
(2,-3,6) . (1,p,-1) = 0
2 + (-3p) + (-6) = 0
-3p - 4 = 0
p = -4/3 [B]

15. a.b.b= 5.3+1.(-1)+7.2 / I akar 3^2+(-1)^2 + 2^2 . b 

= 15 + (-1) + 14 / I akar 9 + 1 + 4 

= 28 / |akar 14|^2 . b 

= 28/14 . 2 . b 

2 (3i-j+2k)

= 6i - 2j + 4k [B]

16. ( 3 -2 1) - ( 2 y 2) / 4+4+y^2 = 1/2 akar 4+4+y^2 

(6-2y+2)2= 8+y^2 

16 - 4y = 8 + y^2

y^2 + 4y -8 =0

y= 4 +- akar 16+32 / 2 = -2+2akar 3 [C]

17. 

18.  maka { -1 <_ x <_ 2}
Maka |p| <_ 2|a| untuk { -1 <_ x <_ 2 } [C]

19. Cari dulu nilai a

u.v = 18

2a-6-8=18

2a=18+6+8

2a=32

a=16

maka u + v = 2i+j+2k + 4i+2j+4k = 6i+3j+6k

20. Pembahasan

A = (4,7,0)

B = (6,10,-6)

C = (1,9,0)

AB = vektor B - vektor A

AB = (6,10,-6) - (4,7,0)

AB = (2,3,-6) > u

AC = vektor C - vektor A

AC = (1,9,0) - (4,7,0)

AC = (-3,2,0) > v

u • v = |u| × |v| × cosA

(2 × -3) + (3 × 2) + (-6 × 0) = √2^2 + 3^2 + (-6)^2 × √(-3)^2 + (2)^2 × cosA

-6 + 6 = 7 × √13 × cosA

0 = 7√13 × cosA

0/7√13 = cosA

0 = cosA

90° = A