sudut antara vektor a= -i+j dan vektor b= i-2j+2k adalah... dikasih penjelasannya

VEKTOR

cari dulu
a•b = (-1, 1, 0) • (1, -2, 2)
a•b = -1 + (-2) + 0
a•b = -3

lalu panjang a dan panjang b
|a| = akar((-1)^2 + 1^2 + 0^2)
|a| = akar(1+1+0)
|a| = akar2

|b| = akar(1^2 + (-2)^2 + 2^2)
|b| = akar(1+4+4)
|b| = akar9
|b| = 3

misalkan sudutnya @

cos@ = (a•b)/|a||b|
cos@ = -3/(akar2 × 3)
cos@ = -1/akar2
cos@ = -1/2 akar2
cos@ = cos135°
@ = 135°