→
AB = [ 2 − 4, 2 − 1] = [ − 2 , 1 ]
→
AC = [3 − 4, 4 − 1] = [ − 1 , 3 ]
więc
iloczyn skalarny
→ →
AB o AC = (−2)*(−1) + 1*3 = 5
Długości wektorów
I AB I = √ 4 + 1 = √5
I AC I = √(−1)2 + 32 = √10
zatem
| 5 | 5 | √2 | ||||
cos α = | = | = U{1}[√2} = | ||||
| √5*√10 | 5*√2 | 2 |