Question

tolong bantu

Diketahui titil P(-3,-1,-5) titik q -1,2,0 dan R 1,2,-2 jika vektor a = vektor PQ dan vektor b = vektor QR + PR tentukan sudut antara vektor a dan b​

Answer 1

Jawab:

Penjelasan dengan langkah-langkah:

PQ = a = (-1 - (-3), 2 - (-1), 0 - (-5))

a = (-1 + 3, 2 + 1, 0 + 5)

a = (2, 3, 5)

QR = (1 - (-1), 2 - 2, -2 - 0)

QR = (1 + 1, 2 - 2, -2 - 0)

QR = (2, 0, -2)

PR = (1 - (-3), 2 - (-1), -2 - (-5))

PR = (1 + 3, 2 + 1, -2 + 5)

PR = (4, 3, 3)

b = (2, 0, -2) + (4, 3, 3)

b = (2 + 4, 0 + 3, -2 + 3)

b = (6, 3, 1)

lal² = 2² + 3² + 5² = 4 + 9 + 25 = 38

lbl² = 6² + 3² + 1² = 36 + 9 + 1 = 46

α = sudut antara a dan b

a.b = (2 x 6) + (3 x 3) + (5 x 1)

a.b = 12 + 9 + 5

a.b = 26

cos α = (a.b)/lallbl

cos α = 26 / √(38 x 46)

cos α = 26/41.8

cos α = 0.287

α = 73°