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) ✔