Given: A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)
To prove: Points A, B and C are collinear
Formula used:
Points A, B and C are collinear if AB + BC = AC or AB + AC = BC or AC + BC = AB
The distance between any two points (a, b, c) and (m, n, o) is given by,
\(\sqrt{(a-m)^2+(b-n)^2+(c-0)^2}\)
Therefore,
Distance between A(4, -3, -1) and B(5, -7, 6) is AB,
= \(\sqrt{(4-5)^2+(-3-(-7)^2+(-1-6)^2}\)
= \(\sqrt{(-1)^2+(-4)^2+(-7)^2}\)
= \(\sqrt{1+16+49}\)
= \(\sqrt{66}\)
Distance between B(5, -7, 6) and C(3, 1, -8) is BC,
= \(\sqrt{(5-3)^2+(-7-1)^2+(6-(-8))^2}\)
= \(\sqrt{(-2)^2+(-8)^2+(14)^2}\)
= \(\sqrt{4+64+196}\)
= \(\sqrt{264}\)
= \(2\sqrt{66}\)
Distance between A(4, -3, -1) and C(3, 1, -8) is AC,
= \(\sqrt{(4-3)^2+(-3-1)^2+(-1-(-8))^2}\)
= \(\sqrt{(1)^2+(-4)^2+(7)^2}\)
= \(\sqrt{1+16+49}\)
= \(\sqrt{66}\)
Clearly,
AB + AC
= \(\sqrt{66}+\sqrt{66}\)
= \(2\sqrt{66}\)
= BC
Hence, A, B and C are collinear
