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Verified answer
1. |a+b|² = |a|²+|b|²+2|a||b| cosα(√5)² = (√2)²+(√9)²+2√2×3 cosα
cosα = -
cosα = -×
cosα = -√2
α = arc cos -
α = 135°
2. ( -1,2,-1)
Verified answer
Mapel : MatematikaKelas : 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