1. Diketahui IaI = √2 , IbI = √9 , Ia + bI = √5 . Besar sudut antara vektor a dan vektor b adalah... 2. Jika w adalah vektor proyeksi orthogonal dari vektor v

1. |a+b|² = |a|²+|b|²+2|a||b| cosα
    (√5)² = (√2)²+(√9)²+2√2×3 cosα
    cosα = -[tex] \frac{2+9-5}{6 \sqrt{2} } [/tex]
    cosα = -[tex] \frac{6}{6 \sqrt{2} } [/tex]×[tex] \frac{ \sqrt{2} }{ \sqrt{2} } [/tex]
    cosα = -[tex] \frac{1}{2} [/tex]√2
    α = arc cos -[tex] \frac{1}{2} \sqrt{2} [/tex]
    α = 135°

2. [tex] \frac{2.-1 + (-3).2 + 4.(-1)}{ \sqrt{(-1) ^{2} + 2 ^{2} + (-1) ^{2} } } [/tex] ( -1,2,-1)
    [tex] \frac{-2-6-4}{ \sqrt{6} ^{2} } (-1,2,-1)[/tex]
    [tex] -\frac{12}{6}(-1,2,-1) [/tex]
    [tex]-2 (-1,2,-1)[/tex]
    [tex](2,-4,2)[/tex]

Mapel : Matematika
Kelas : X SMA
Bab : Vektor

Pembahasan :
1||

|a + b| = √5
|a + b|² = 5
|a|² + |b|² + 2|a| |b| Cos β = 5
(√2)² + (√9)² + 2(√2)(√9) Cos β = 5
2 + 9 + 6√2 Cos β = 5
6√2 Cos β = -6
Cos β = -1/2 √2
Cos β = Cos 135°
β = 135°

2||

W = (u . v)/|u|² . u
W = (-2 - 6 - 4)/6 [-1 2 -1]
W = -12/6 [-1 2 -1]
W = -2 [-1 2 -1]
W = [2 -4 2]
W = 2i - 4j + 2k