Jika OA = i + k , OB = j + k dan OC = cj + k dan sudut ABC = 60 derajat. maka c = ...
arsetpopeye
OA = a = i + k = (1 , 0 , 1) OB = b = j + k = (0 , 1 , 1) OC = c = cj + k = (0 , c , 1) sudut ABC = 60° = sudut antara vektor BA dan BC BA = a - b = (1 , -1 , 0) => |BA| = √(1^2 + (-1)^2 + 0^2) = √2 BC = c - b = (0 , c - 1, 0) => |BC| = √(0^2 + (c - 1)^2 + 0^2) = √(c^2 - 2c + 1) BA . BC = 1(0) + (-1)(c - 1) + 0 = 1 - c
OB = b = j + k = (0 , 1 , 1)
OC = c = cj + k = (0 , c , 1)
sudut ABC = 60° = sudut antara vektor BA dan BC
BA = a - b = (1 , -1 , 0) => |BA| = √(1^2 + (-1)^2 + 0^2) = √2
BC = c - b = (0 , c - 1, 0) => |BC| = √(0^2 + (c - 1)^2 + 0^2) = √(c^2 - 2c + 1)
BA . BC = 1(0) + (-1)(c - 1) + 0 = 1 - c
BA . BC = |BA|.|BC| cos 60°
=> 1 - c = √2 . √(c^2 - 2c + 1) . 1/2
=> 2 - 2c = √(2c^2 - 4c + 2)........ kuadratkan
=> 4 - 8c + 4c^2 = 2c^2 - 4c + 2
=> 2c^2 - 4c + 2 = 0
=> c^2 - 2c + 1 = 0
=> (c - 1)(c - 1) = 0
=> c = 1