Jawaban:
Penjelasan dengan langkah-langkah:
u = (2, 3, –1)
v = (–4, x, 2)
p.s.o , u → v = ⅓
u.v = 2(–4) + 3x + (–1)2
u.v = –8 + 3x – 2
u.v = 3x – 10
|v| = √((–4)²+ x² + 2²)
|v| = √(x² + 20)
p.s.o = u.v
|v|
⅓ = 3x – 10
√(x²+20)
••• dikali silang •••
1( √(x²+20) ) = 3(3x – 10)
√(x²+20) = 9x – 30
(x²+20) = (9x – 30)²
x² + 20 = 81x² – 540x + 900
80x² – 540x + 880 = 0
4x² – 27x + 44 = 0
(4x – 11)(x – 4) = 0
4x = 11
x = 11 → tidak bulat
4
x – 4 = 0
x = 4 → bilangan bulat
jadi, vektor v = ( –4, 4, 2 )
nilai dari u + 2v
= (2, 3, –1) + 2( –4, 4, 2 )
= (2–8, 3+8, –1+4)
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Jawaban:
D
Penjelasan dengan langkah-langkah:
V E K T O R
u = (2, 3, –1)
v = (–4, x, 2)
p.s.o , u → v = ⅓
u.v = 2(–4) + 3x + (–1)2
u.v = –8 + 3x – 2
u.v = 3x – 10
|v| = √((–4)²+ x² + 2²)
|v| = √(x² + 20)
p.s.o = u.v
|v|
⅓ = 3x – 10
√(x²+20)
••• dikali silang •••
1( √(x²+20) ) = 3(3x – 10)
√(x²+20) = 9x – 30
(x²+20) = (9x – 30)²
x² + 20 = 81x² – 540x + 900
80x² – 540x + 880 = 0
4x² – 27x + 44 = 0
(4x – 11)(x – 4) = 0
4x = 11
x = 11 → tidak bulat
4
x – 4 = 0
x = 4 → bilangan bulat
jadi, vektor v = ( –4, 4, 2 )
nilai dari u + 2v
= (2, 3, –1) + 2( –4, 4, 2 )
= (2–8, 3+8, –1+4)
= (–6, 11, –3) ✔