Show that the points A(1, -2, -8) B (5, 0, -2) and C(11, 3, 7) are collinear and find the ratio in which B divides AC.

Explanation

The given point are `A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7)`. Therefore,
`vec(AB) = (5-1)hati + (0+2)hatj + (-2 + 8)hatk`
`" " = 4hati + 2hatj + 6hatk`
`vec(BC) = (11 -5)hati + (3-0)hatj + (7+2)hatk`
`= 6hati + 3hatj + 9hatk`
`vec(AC) = (11-1)hati + (3+1)hatj + (7+8)hatk`
`" " = 10hati + 5hatj + 15hatk`
`" "|vec(AB)| = sqrt(4^(2) + 2^(2) + 6^(2))= sqrt(16 + 4+ 36)`
`" " = sqrt(56) = 2 sqrt(14)`
`|vec(BC)| = sqrt(6^(2)+3^(2)+9^(2)) = sqrt(36 + 9 + 81)`
`" "= sqrt(126) = 3sqrt(14)`
`" "|vec(AC)| = sqrt(10^(2) + 5^(2) + 15^(2)) = sqrt(100 + 25 + 225)`
`" " = sqrt(350) = 5sqrt(14)`
`therefore " "|vec(AC)| = |vec(AB)| + |vec(BC)|`
Thus, the given points A, B, and C are collinear,
Also `2 |vec(BC)| = 3|vec(AB)|`
Hence, point B divides AC in the ratio `2:3`.