jika vektor OA=i + k, vektor OB = j + k, dan vektor OC =cj +4k serta sudut ABC =60 derajat, nilai c adalah

vektor
A(1,0,1) , B(0,1,1), C(0,c,4)

<ABC = <(BC,BA) = 60°
cos <ABC = 1/2

BC = c-b = (0, c-1, 3) --> |BC| = √(0+(c-1)²+9) = √(9+(c-1)²)
BA = a - b = (1, -1, 0)  ->|BA| = √(1+1+0) = √2
.
BA.BC = 1(0) + (-1)(c-1)+(0)(3) =  1-c

cos (<ABC) = BA.BC / |BA|BC|
BA.BC = |BA||BC|. cos <ABC
1-c = (√2)(√(9+(c-1)²) (1/2)
2(1-c) = √{2(9 +(c-1)²)}
(2-2c)² = 2 ( 9 + (c-1)²)
(2-2c)² = 18 + 2(c-1)²
(2 -2c)² - 2 (c-1)² = 18
4 - 8c + 4c² - 2c² + 4c - 2 = 18
2c² - 4c +2 - 18 = 0
c² - 2c - 8 = 0
(c - 4)(c+2)= 0
c = 4 atau c = -2