" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
VEKTOR∆PQR
P( 0 , 1 , 4 )
Q( 2 , -3 , 2 )
R( -1 , 0 , 2 )
∠PRQ = ___?
RP = (1 1 2)
|RP| = √(1+1+2²) = √6
RQ = (3 -3 0)
|RQ| √(3²+(-3)²+0) = √18
RP • RQ = |RP| |RQ| cos ∠PRQ
(1 1 2)' (3 -3 0) = √6 • √18 cos ∠PRQ
3 - 3 + 0 = 6√3 cos ∠PRQ
0 = cos ∠PRQ
∠PRQ = 90° ✔️
cek GEOMETRI nya