Jawaban PTS
1) jika vektor a = (1, 2, 3 ) b= ( 5 ,4 ,-1) dan c = (4 ,-1, 1) maka hasil dari operasi vektor a+2b-3c...
: a + 2b -3c = (1,2,3) + 2(5,4,-1) -3(4,-1,1)
= (-1,13,-2) (d)
2) diketahui |a| =√3 , |b| =1 dan |a-b|= 1 panjang vektor a+b adalah...
: |a+b| =2(|a| + |b|²) - |a-b|²
= 2[(√3)²+1²]-1²
= 2 . (3+1)-1 =7
|a+b| =√7 (c)
3)diketahui a = 2i - 3j + 4k dan b = 5j +5k. Nilai a.b adalah
: a.b = (2i-3j+4k) . (0i+5j+5k)
= (2i.0i)-(3j.5j)+(4k.5k)
= -15j +20k
= -15+20
= 5 (b)
4) diketahui |a-b| = 2√19 jika |a| =4 dan |b| =6 maka |a+b| ...
: (a+b) = (a)²+(b)² + 2a.b
= |a|² + |b|² + 2 |a| |b| cos ∅
(2√19)² =4²+6²2.(4)(6) cos ∅
Cos ∅ =1/2
∅ =60°
(a-b)² = (a)²+(b)²-2a.b
= 4²+6²-2(4)(6)(1/2)
(a-b) =√28
= 2√7 (a)
5) diketahui vektor a = 2i - 3j +k , b =pi+2j-k dan c = i-j+3k. Jika b tegak lurus terhadap vektor C maka vektor a-b-c...
: 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) jika sudut antar vektor a = i + √2 j + p k dan vektor b = i-√2 j + p k adalah 60° maka p adalah..
: a.b = |a| |b| cos 60°
( 1 √2 p ) (1-√2p) = √1²+(√2)²+p² √1²+(-√2)²+p² (1/2)
1-2 +p = √1+2+p² √1+2+p² (1/2)
2(-1+p²) = 1+2+p²
-2+2p² = 3+p²
P=3+2
P ±√5 (d)
7) titik A(3,2,-1) B(1,-2,1) dan C(7,p,-1,-5) segaris untuk nilai p
: (7 , p-1 ,-5) - (3,2,-1) =k [(1,-2,1) - (3,2,-1)]
= (4, p-3_-4)= k (-2,-4,2)
maka berlaku 4= k.(-2) = k=-2
P-3=k (-4) = (-2)(-4) = 8
= p-3 =8
p = 8+3
p = 11 (d)
8) : titik P = 3b+2a / 3+2
= 3(3,-4,6) + 2(3,1,-4) / 5
= (9,-12,18)+(6,2,-8) / 5
= (15,-10,10) / 5 = (3,-2,2)
Sehingga
PC = PB+BC = (b-p) + (c-b)
= ([3,-4,6] - [3,-2,2]) + ([-1,5,4]-[3,-4,6])
= [0,-2,4] + [-4,9,-2] = [-4,7,2] (e)
9) : menggunakan rumus proyeksi orthagonal
a.b / 1b1 = 8
8p + 16 √p²+16 = 8
8p +16 = 8√p²+16
( 8p+16)² = (8√p²+16)²
64p²+256 p + 256 = 64 p² +1.024
256 p = 1.024 -256
256 p = 768
P = 768 / 256
P = 3 (c)
10) a.b =0 (p,2,-1) . (4,-3,6)
= 0 = 4p +2(-3)-1.6
P = 3
( a-2b).(3c) =[ (3,2,-1) -2(4,-3,6). (3(2,-1,3)]
= (-5,8,-13). (6,-3,9)
= -5.6+8(-3)+(-13).9
= -171 (e)
11) AB =√(3-1)²+(3-2)²+(1-3)²
= √2²+1²+(-2)²
= √4+1+4
=√9 =3
BC = √(7-3)²+(5-3)²+(-3-1)²
=√4²+2²+(-4)²
= √36 = 6
AB / BC = 3/6 =½
AB : BC = 1:2 (A)
12) |a+b| = |a-b|
= |a|²+|b|² + 2a.b =|a|² + |b|² - 2a.b
= 4a.b = 0 =a.b =0
Sehingga vektor a dan b saling tegak lurus (a)
13) c = [ u.v / |v|²] v = [-42/14] (1,2,3
= -3 (1,2,3) = (-3,-6,-9)
= -3i-6j-9k (a)
14) (2,-3,6) . (1,p,-1) = 0
2+(-3p) + (-6) = 0
-3p -4 =0
P = -4/3 (b)
15) (a.b/b.b). b = (15-1+14/9+1+4) . 3i-j+2k
= (2). (3i-j+2k)
= 6i-2j+4k (b)
16) (3,-2,1) (2,y,2) / 4+4+y² = ½ √4+4+y²
(6-2y+2)2=8+y²
16-4y=8+y²
Y²+4y-8=0
y = 4±√16+32 / 2 = -2 +2√3 (c)
17) 2√2 (c)
18) |p| = a.b / |a| = (x+1 , x ) . (2x ,3x +1) / √(x+1)²+x² = |p|=5x²+3x / √2x²+2x+1
|p| ≤ 2|a|
= 5x²+3x / √2x²+2x+1 ≤ 2 √2x²+2x+1
= 5x²+3x ≤ 4x² + 4x + 2
= x²-x-2 ≤ 0
= (x-2) (x+1) = 0
-1 ≤ x ≤ 2 (c)
19) vektor u= 2i+j+k dan v =4i+2j+4k adalah vektor searah
Nilai u.v = 18
: (2.4) + (1.2) + (2.4)
8 + 2 + 8 = 18
18 = 18
Petunjuknya : a) pernyataan benar,alasaan benar, dan mempunyai hubungan sebab akibat
20) AB = B-A
(6-4) (10-7) + (-6-0)
2,3,-6
AC = C-A
(1-4) (9-7) (0-0)
-3,2,0
AB.AC = (2.-3) + (3.2) + (-6.0)
-6+6+0
0
AB.AC > 0
0 > 0
Pernyataan yang tepat adalah :
D) pernyataan salah alasan benar
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