Nama : REGITAkelas : X MIPA 3
absen : 33
Pembahasan soal PAT kelas X semester 2
Question 1
1. Diketahui :
a = i - 8j + 5k, b = 3i + 8j + 2k, c = -2i - 4j + 3k
Ditanya :
a + 2b - 3c = .....?
Jawab :
i - 8j + 5k + 2 (3i + 8j + 2k) - 3(- 2i - 4j + 3k)
i -8j + 5k + 6i + 16j + 4k + 6i + 12j - 9k
13i + 20j
Jadi,
a + 2b - 3c adalah 13i + 20j
2. Pembahasan :
Diketahui :
a = (7, -4 ,p), b = (1, 8, -5), c = (1, -2, -3)
Vektor b tegak lurus vektor a
Ditanya :
a - b - c = ....?
Jawab :
Vektor b tegak lurus vektor a ➡️ b.a = 0
(1, 8, -5) . (7, -4, p) = 0
1(7) + 8(-4) + -5(p) = 0
7+ (-32) + (-5p) = 0
-25 + (-5p) = 0
-5p = 25
p = -5
a - b - c =....
(7, -4, -5) - (1, 8, -5) - (1, -2, -3)
(7 - 1 -1),(-4 - 8 + 2),(-5 + 5 + 3)
(5, -10, 3)
Jadi a - b - c yaitu (5, -10, 3)
3. Pembahasan :
Diketahui :
Xp = 11, Yp = 9
Xq = -4 , Yq = 1
PQ adalah sebuah vektor dengan titik pangkal P dan titik ujung Q
Ditanya :
Panjang vektor PQ =....?
Jawab :
PQ = (Xq - Xp, Yq - Yp)
PQ = (-4 - 11, 1- 9)
PQ = (-15, -8)
panjang vektor PQ =|PQ|
|PQ| = √(-15)^2 + (-8)^2
|PQ| = √225 + 64
|PQ| = √289
|PQ| = 17
Jadi,panjang vektor PQ yaitu 17
Question 2
4. Pembahasan :
Diketahui :
A = (1, -5, -9), B = (7, -14, 6), C = (15, -26, 20)
A, B, C segaris (kolinear)
Ditanya :
AB : BC =.....?
Jawab :
Rumus : AB =kBC
B - A = k (C - B)
(7, -14, 6) - (1, -5, -9) = k[(15, -26, 20) - (7, -14, 6)]
(7-1, -14+5, 6+9) = k(15-7, -26 + 14, 20-6)
(6, -9, 15) = k(8, -12, 14)
6 = 8k
k = 6/8 = 3/4
Jadi,perbandingan AB : BC adalah 3 : 4
5. Pembahasan :
Diketahui :
A = (-3, 5, 2),B = (2,-1,8), C= (3, -4, 6)
AB = u, BC = v
Ditanya :
u.v =...?
Jawab :
AB = B - A = u
(2,-1,8) - (-3,5,2) = (2+3,-1-5,8-2) = (5, -6, 6)
BC = C - B =v
(3, -4, 6) - (2, -1,8) = (3-2, -4+1, 6-8) = (1, -3, -2)
u.v =...
(5,-6,6) . (1, -3, -3) = 5(1), -6(-3), 6(-3) = (5,18,-12)
Atau 5 + 18 - 12 = 11
Jadi u.v adalah (5,18,-12) atau 11
6. Pembahasan :
Diketahui :
P=(-8,-11,-15), Q=(-11,20,10), R=(12,22,-20)
PQ=a, QR +PR = b
Ditanya :
a . b =....?
Jawab :
PQ= Q - P = a
(-11,20,10) - (-8,-11,-15)
= (-11 + 8, 20+11, 10+15)
= (-3,31,25)
QR + PR = (R - Q) + (R - P) = b
(12,22,-20) - (-11,20,10) + (12,22,-20) - (-8,-11,-15)
(12+11, 22-20, -20-10) + (12+8, 22-11, -20+15)
(23, 2, -30) + (20,11, -5) = (23+20,2+11, -30-5) = (43,13,-35)
a . b =....
(-3,31,25) . (43,13,-35) = -3 (43), 31(13),25(-35) = (-129, 403, -875)
Atau
-129 + 401 - 875 = -605
Jadi a . b yaitu (-129, 403,-875) atau -605
7. Pembahasan :
diektahui :
P = (-1, 10), Q = (5, 3), R = (3, 15)
PR = a, PQ = b, QR = c
c = ma + nb
ditanya :
m + n = ...?
jawab :
c = ma + nb
QR = m(PR) + n(PQ)
R - Q = m(R - P) + n(Q - P)
(3, 15) - (5, 3) = m[(3, 15) - (-1, 10)] + n[(5, 3) - (-1, 10)]
(3-5, 15-3) = m(3+1, 15-10) + n(5+1, 3-10)
(-2, 12) = m(4, 5) + n(6, -7)
(-2, 12) = (4m, 5m) + (6n, -7n)
4m + 6n = -2 .... (1)
5m - 7n = 12 ... (2)
eliminasi
4m + 6n = -2 |x5|
5m - 7n = 12 |x4|
20m + 30n = -10
20m - 28n = 48 -
58n = -58
n = -58/58 = -1
subtitusikan n = -1 ke persamaan (1)
4m + 6n = -2
4m + 6(-1) = -2
4m - 6 = -2
4m = -2 + 6
4m = 4
m = 4/4 = 1
m + n = 1 + (-1) = 0
jadi m + n adalah 0
8.Pembahasan :
diketahui :
u = 5i - 6j + 2k, v = i + pj - 4k, w = -3i + 4j - 5k
u tegak lurus dengan v
ditanya :
nilai p = ...?
jawab :
u tegak lurus v ➝ u . v = 0
(5 - 6 + 2) . (1 + p - 4) = 0
5(1), -6(p), 2(-4) = 0
5, -6p, -8 = 0
-3 = 6p
p = -3/6 = -1/2
jadi nilai p adalah -1/2
9.Pembahasan :
diketahui :
a = 5i + 6j - 2k, b = 3i - 4j - 13k, c = a - b
ditanya :
vektor satuan yang searah dengan c = ....?
jawab :
c = a - b = (5i + 6j - 2k) - (3i - 4j - 13k) = 5i-3i, 6j+4j, -2k+13k = 2i, 10j, 11k
|C| = √2²+10²+11² = √4+100+121 = √225 = 15
c = (2, 10, 11) / 15 atau c = 2/15i + 10/15j + 11/15k
jadi vektor satuan yang searah dengan c adalah 2/15i + 10/15j + 11/15k
10. Pembahan
diketahui :
P = (-3, -6, 9), Q = (6, 0, 12), R = (3, -9, 18)
a = PQ, b = QR
ditanya :
besar sudut antara a dan b = ...?
jawab :
a = PQ = Q - P = (6, 0, 12) - (-3, -6, 9) = (6+3, 0+6, 12-9) = (9, 6, 3)
b = QR = R - Q = (3, -9, 18) - (6, 0, 12) = (3-6, -9-0, 18-12) = (-3, -9, 6)
a . b = (9, 6, 3) . (-3, -9, 6) = 9(-3), 6(-9), 3(6) = (-27, -54, 18) atau -27 - 54 + 18 = -63
|a| = √9²+6²+3² = √81+36+9 = √126 = 3√14
|b| = √(-3)²+(-9)²+6² = √9+81+36 = √126 = 3√14
cos α = (a.b)/(|a||b|) = -63/(3√14.3√14) = -63/3.14.3 = -7/14 = -1/2
α = -1/2 maka cos α 120⁰ atau 240⁰
jadi besar sudut antara a dan b adalah 120⁰ atau 240⁰
11.Pembahasan :
diketahui :
u = 3i - 3j, v = 2i - 4j - 2k, w = i + j + 2k
AB = u, BC = v, CA = w
ditanya :
besar sudut ACB = ...?
jawab :
cari nilai vektor CA dan CB
CA = w = i + j + 2k = 1 + 1 + 2
CB = BC = v = 2i - 4j - 2k = 2 - 4 - 2
CA . CB = (i + j + 2k) . (2i - 4j - 2k) = 1(2) + 1(-4) + 2(-2) = 2 - 4 - 4 = -6
|CA| = √1²+1²+2² =√1+1+4 = √6
|CB| = √2² +(-4)²+(-2)² = √4+16+4 = √24 = 2√6
cos α = (CA .CB)/(|CA||CB|) = -6/(√6 . 2√6) = -6/(2 . 6) = -6/12 = -1/2
α = -1/2 maka cos α 120⁰ atau 240⁰
jadi besar sudut antara ACB adalah 120⁰ atau 240⁰
12.Pembahasan :
diketahui :
g melalui : A = (3, -9, -5), B = (5, -8, -8)
h melalui : C = (11, -12, -2), D = (10, -9, -4)
ditanya :
besar sudut g dan h = ...?
jawab :
g = AB = B - A = (5, -8, -8) - (3, -9, -5) = (5-3, -8+9, -8+5) = (2, 1, -3)
h = CD = D - C = (10, -9, -4) - (11, -12, -2) = (10-11, -9+12, -4+2) = (1, 3, -2)
g . h = (2, 1, -3) . (1, 3, -2) = 2(1) + 1(3) + -3(-2) = 2 + 3 + 6 = 11
|g| = √2²+1²+(-3)² = √4+1+9 = √14
|h| = √1²+3²+(-2)² = √1+9+4 = √14
cos α = (g .h)/(|g||h|) = 11/(√14 . √14) = 11/14 = 0,785
α = 0,785 maka cos α 38⁰ atau ±360⁰
jadi besar sudut g dan h adalah 38⁰ atau 322⁰
13.pembahasan :
diketahui :
|a| = √7, |b| = 2√3, |a - b| = 3√3
ditanya :
panjang vektor a + b = ...?
jawab :
|a - b|² = a² - 2abcosθ + b²
(3√3)² = (√7)² - 2ab + (2√3)²
27 = 7 - 2ab + 12
2ab = 7 + 12 - 27
2ab = 19 - 27
2ab = -8
ab = -8/2 = -4
|a + b|² = a² + 2ab² + b²
|a + b|² = (√7)² + 2(-4) + (2√3)²
|a + b|² = 7 - 8 + 12
|a + b|² = 11
|a + b| = √11
jadi panjang vektor a + b adalah √11
14.Pembahasan :
diketahui :
A = (8, -11, 3), vektor AB = (7, 4, -9)
ditanya :
koordinat B = ....?
jawab :
AB = B - A
(7, 4, -9) = B - (8, -11, 3)
B = (7, 4, -9) + (8, -11, 3)
B = (15, -7, -6)
jadi koordinat B adalah (15, -7, -6)
15.Pembahasan :
diketahui :
u = 5i + 3j - 3k, v = 9i + 3j - 3k, w = 13i + 7j - 3k
a = v - u, b = w - v
a dan b diapit sudut θ
ditanya :
tan θ = ...?
jawab :
a = v - u = (9 + 3 - 3) - (5 + 3 - 3) = (9-5, 3-3, -3+3)
a = (4, 0, 0)
b = w - v = (13 + 7 - 3) - (9 + 3 - 3) = (13-9, 7-3, -3+3)
b = (4, 4, 0)
a . b = (4, 0, 0) . (4, 4, 0) = 4(4) + 0(4) + 0(0) = 16
|a| = √4²+0²+0² = √16 = 4
|b| = √4²+4²+0² = √32 = 4√2
cos θ = (a.b)/(|a||b|) = 16/(4 . 4√2) = 16/16√2 = 1/√2 = 1/2√2
cos θ = 1/2√2 = 45°
tan θ = tan 45° = 1
jadi nilai tan θ adalah 1
16.Pembahasan :
diketahui :
u = (-1, 4, 6), v = (-5, 3, p)
proyeksi skalar orthogonal u pada arah v = 1/2 v
ditanya :
nilai p = ....?
jawab :
u . v = (-1, 4, 6) . (-5, 3, p) = -1(-5) + 4(3) + 6(p) = 5 + 12 + 6p = 17 + 6p
|v| = √(-5)²+3²+p² = √25+9+p² = √34+p²
(u . v)/|v| = 1/2 . |v|
(17 + 6p)/(√34+p²) = 1/2 . √34+p²
(17 + 6p)/(√34+p²) = (√34+p²)/2
2(17 + 6p) = √34+p² . √34+p²
2(17 + 6p) = 34 + p²
2(17 + 6p) - 34 - p² = 0
34 + 12p - 34 - p² = 0
12p - p² = 0
p(12 - p) = 0
p = 0 atau 12 - p = 0 ➝ p =12
jadi nilai p adalah p = 0 atau p = 12
17.Pembahasan :
diketahui :
a = -4i + j + 2k, b = 4i - 5j + 3k
ditanya :
proyeksi orthogonal a pada b = ....?
jawab :
ab = (a . b)/(|b|)² . b
ab = (-4.4 + 1.-5 + 2.3)/(√4² + (-5)² + 3²)² . b
ab = (-16 - 5 + 6)/(√16 + 25 + 9)² . b
ab = -15/50 . b
ab = -3/10 (4i - 5j + 3k)
ab = -6/5i + 3/2j + 9/10k
ab = (-6/5, 3/2, 9/10)
jadi proyeksi orthogonal a pada b adalah (-6/5, 3/2, 9/10)
Question 3
18). 5^2x+1 > 5^x+4
(5^x)² × 5¹ > 5^x + 4
5 p² > p + 4
5p² - p - 4 > 0
(p - 1) (5p + 4) > 0
p = 1 p = -4/5
HP : {x < -4/5 atau x > 1}
19). 2^x × 2¹ × 2^-4 ≤ 0 / 1-2^x
p - 2p - 4 / 1-2x ≤ 0
(p-2) (p+1) / p (1-p) ≤ 0
p = 2 p = -1
HP : { -1 < x < 2}
20). (4^x)² × 4¹ > 4^x + 3
4p² - p - 3 > 0
(4p - 4/4) (4p +3) > 0
(p - 1) (4p + 3) > 0
p = 1 p = -3/4
HP : {-3/4 < x < 1}
21). 3^x-2y = 3^-4
x - 2y = -4
x - y = 4
__________________ -
-y = 8
y = -8
x-8 = 4
x = 12
x+y = 12 + (-8) = 4
22). (2a^5 - b^-5)/3 × 2a^9 × b^-1) ^-1
(2¹a^5 b^-5/2^5 a^9 b^-1)
(2^-4 a^-4 b^-4)
(2^⁴ a^⁴ b^⁴)
(2 ab)⁴
23). 9^3x-4 = 81^½x-5
(3²)^3x-4 = (3^-4)^2x-5
6x - 8 = -8x + 20
14x = 28
x = 2
24). 4¹+² × 3^4x+1 < 432
4 × 4^2x × 3^4x × 3 < 432
4^2x × 3^4x < 36
16^x × 18^x < 36
1.296^x < 36
36^x < 36
x < 1
25). (1/3)^x+2 < (1/3)^x
x + 2 < x
0 < 2 (TIDAK MEMENUHI)
(1/3)^x+2 < (1/3)^-x
x +2 < -x
2x < -2
x < -1
Question 4
32. g = ( 2, 1, -3 ) h = ( -1, 3, -2 )
ǀ g ǀ = √14 ǀ h ǀ = √14
Cos Ɵ = g.h / ǀ g ǀ. ǀ h ǀ
= -2 + 3+6 / √14. √14
= 7/14
= 1/2
33. ǀuǀ . ǀv ǀ + ǀu-vǀ = ǀu + v ǀ
45 + 4√5 = ǀu + v ǀ
ǀu + v ǀ = 53,944 = 54
34. 34. AB = B – A
AB + A = B
(-15, 12 ) + (-4, -6 ) = B
(-19, 6 ) = B
35. C = u.v / ǀvǀ . v
= -6 + 2 + (-3) / (√14)2 . ( -3i + 2j – k )
= -7/14 . ( -3i + 2j – k )
= -1/2 . ( -3i + 2j – k )
= ( 3/2i -j + 1/236 4
= 14 + 90 -11m / √4+100+121
4 = 104 – 11m / √225
4 = 104 – 11m / 15
60 = 104 – 11m
11m = 104 – 60
11m = 44
m = 4k
37. p . q = | p | | q | cos Θ
p . q = (-3i - 3aj - bk ) . ( -ai + bj - 2ak )
p . q = 3a - 3ab + 2ab
p . q = 3a
| p |² = (-3)² + (-3a)² + (-b)² = 9 + 9a² + b²
| p | = √ (9 + 9a² + b² )
| q |² = (-a)² + b² + (-2a)² = a² + b² + 4a²
| q | = √ (a² + b² + 4a² ) = √(5a²+b²)
r = [ ( p . q ) / (| q |²) . q
i - j + 2k = [ 3a /(5a²+b²) . ( -ai + bj - 2ak )
-3 = 3a² /( 5a² + b² )
-3a² - b² = 3a²
-b² = a²
cos Θ = p . q / ( | p | | q | )
(√6)/8 = 3a / ( √ (9 + 9a² + b² ) . √(5a²+b²)
[ (√6)/8 ]² = (-3b)² / (9 - 9b² + b² ) . (-5b²+b²)
( 6 / 64 ) (9 - 8b²) . (-4b²) = 9b²
( -24 / 64 ) (9 - 8b²) = 9
-8b²+ 9 = -24
-8b² = -33
b² = 4,125
b = √4,125
b = 2,03
maka, b = -a = -2,03. Jadi, nilai a = -2,03
38). Cos x = a.b/|a|.|b|
½√2 = -12 + 2x / √20 . √9+x²
½√2 = -12 + 2x / √180 + 20x²
½√2 . √180 + 2x² = -12 + 2x
½ √360 + 40x² = -12 + 2x
(√360 + 40x²)² = (-24 + 4x)²
360 + 40x² = 16x² - 192 + 576
24x² + 192x - 216 = 0
x² + 8x - 9 = 0
(x+9) (x-1) = 0
x = -9 x = 1
Jadi, HP nya bilangan yang positif yaitu x = 1
39). |r| = p.q / |p|
2 = 6 + b + 2 / 3
12 = 8 + b
4 = b
Jadi, b = 4
40). AB = (5,1,1)
BC = (-3, -2, 2)
D = AB.BC / |BC| ×BC
= -15 + (-2) + (-2) / (√9+4+4)² × (-3,-2,2)
= -19/17 (-3, -2, 2)
= 57/17, 38/17, -38/17
Jadi, HP = {57/17, 38/17, -38/17}
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