Diketahui D(3,2,-1), E(2,1,0), dan F(-1,2,3), Kosinus sudut antara ruas garis berarah DE dan EFadalah.

D(3, 2, -1)
E(2, 1, 0)
F(-1, 2, 3)

Kosinus DE dan EF

DE = e - d = (2, 1, 0) - (3, 2, -1) = (-1, -1, 1)
EF = f - e = (-1, 2, 3) - (2, 1, 0) = (-3, 1, 3)

Cos Φ = DE. EF / |DE| . |EF|

DE. EF = (-1, -1, 1) . (-3, 1, 3) = (3 - 1 + 3) = 5
|DE| = √(-1)² + (-1)² + (1)² = √3
|EF| = √(-3)² + (1)² + (3)² = √20 = 2√5

Cos Φ = (5) / (√3 × 2√5)
cos Φ = 5 / 2√15
Cos Φ = (1/6) √15

Maka kosinusnya 1/6. √15