Question

Diketahui ∆ ABC dengan titik
A(2, –1, – 3), B(–1, 1, –11), dan C(4, –3, –2). Proyeksi vektor AB pada AC adalah …

a. –12i + 12j – 6k
b. –6i + 4j – 16k
c. –4i + 4j – 2k
d. –6i – 4j + 16k
e. 12i – 12j + 6k

@ Wello1
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Answer 1

AB = B - A = (-1, 1, -11) - (2, -1, -3) = (-3, 2, -8)
AC = C - A = (4, -3, -2) - (2, -1, -3) = (2, -2, 1)

misal AB = u dan AC = v

proyesi u pada v = (u.v)/||v||² * v

u.v = (-3, 2, -8).(2, -2, 1) = -6 - 4 - 8 = -18
||v||² = (√(2² + (-2)² + 1²))² = 2² + (-2)² + 1 = 9

u.v/||v||² = -18/9 = -2

u.v/||v||² * v = -2 * (2, -2, 1)
= (-4, 4, -2)
= -4i + 4j - 2k

Answer 2

AB = vektor B - vektor A
= (-1, 1, -11) - (2, -1, -3)
= (-3, 2, -8)
AC = vektor C - vektor A
= (4, -3, -2) - (2, -1, -3)
= (2, -2, 1)

proyeksi vektor = [(vektor AB × vektor AC)/ pjg vektor AC^2] × vektor AC
= [(-3, 2, -8)(2, -2, 1) / (√(2^2 + (-2)^2 + 1^2))^2] × (2, -2, 1)
= [(-6 - 4 - 8) / (4 + 4 + 1)] × (2, -2, 1)
= [(-18) / 9] × (2, -2, 1)
= -2(2, -2, 1)
= (-4, 4, -2)
= -4i + 4j - 2k (c)