Jawaban:
C.
Penjelasan dengan langkah-langkah:
PERBANDINGAN VEKTOR
PQ : PR = 1 : 3
QP : PR = –1 : 3
–PR = 3QP
P(2, 1, –2)
Q(0, 2, –1)
xR = 3(0) – 2 = –2 = –1
3–1 2
yR = 3(2) – 1 = 5
zR = 3(–1) + 2 = –1
xR = –1, yR = 5/2, zR = –½
koordinat titik R
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Jawaban:
C.
Penjelasan dengan langkah-langkah:
PERBANDINGAN VEKTOR
PQ : PR = 1 : 3
QP : PR = –1 : 3
–PR = 3QP
P(2, 1, –2)
Q(0, 2, –1)
xR = 3(0) – 2 = –2 = –1
3–1 2
yR = 3(2) – 1 = 5
3–1 2
zR = 3(–1) + 2 = –1
3–1 2
xR = –1, yR = 5/2, zR = –½
koordinat titik R
= { –1, 5 , –½ } ✔
2