Show that the points `A(1,-2,-8),B(5,0,-2)a n dC(1,3,7)` are collinear, and find the ratio in which `B` divides `A Cdot`

Explanation

`vec(A)B` = Position vector of B - position vector of A
`=(5hat(i) + 0hat(j) - 2hat(k)) - (hat(i) - 2 hat(j) - 8hat(k))`
`=4hat(i) + 2hat(j) + 6hat(k)`
`|vec(AB) = sqrt(4^(2) + 2^(2) + 6^(2))`
`= sqrt(16 + 4 + 36) = sqrt(56) = 2sqrt(14)`
`vec(BC)` = Position vector of C - Position vector of B
`=(11hat(i) + 3hat(j) + 7hat(k)) - (5hat(i) + 0hat(j) - 2hat(k))`
` = 6hat(i) + 3hat(j) + 9hat(k)`
`|vec(BC)|=sqrt(6^(2) + 3^(2) + 9^(2))`
`= sqrt(36 + 9 + 81) = sqrt(126) = 3sqrt(14)`
`vec(AC)` = Position vector of C - Position vector of A
`=(11hat(i) + 3hat(j) + 7hat(k)) - (hat(i) - 2hat(j) - 8 hatk)`
`=10hat(i) + 5hat(j) + 15hat(k)`
`|vec(AC)| = sqrt(10^(2)+5^(2)+15^(2))`
`=sqrt(100+25+225)=sqrt(350) = 5sqrt(14)`
`therefore |vec(AC)|=|vec(AB)|+|vec(BC)|`
Therefore, the given points A, B and C collinear. Let point B is on the line AC such that it divides AC in the ratio `lambda : 1`.
Position vector of B
`=(lambdaxx "Position vector of C + 1"xx "Position vector of A")/(lambda+1)`
`=(1)/(lambda + 1){lambda(11hat(i) + 3hat(j)+ 7hat(k))+1(hat(i) - 2hat(j) - 8hat(k))}`
`=((11lambda + 1)/(lambda + 1))hat(i) + ((3lambda - 2)/(lambda + 1))hat(j) + (7lambda -8)/(lambda+1) hat(k)`
`implies((11lambda + 1)/(lambda + 1))hat(i) + ((3lambda - 2)/(lambda + 1))hat(j) + (7lambda -8)/(lambda+1) hat(k)`
`=5hat(i) + 0hat(j) - 2hat(k)`
`implies (11lambda + 1)/(lambda + 1) = 5, (3lambda - 2)/(lambda + 1) = 0 " and "(7lambda-8)/(lambda+1) =-2`
`implies 11lambda + 1=5lambda + 5, 3lambda = 2, 7lambda - 8 =- 2lambda - 2`
`implies 6lambda = 4,lambda = (2)/(3), 9lambda = 6 implies lambda = (2)/(3)`
Therefore, points A, B and C are collinear and B divides AC in the ration `(2)/(3) : 1` i.e., 2 : 3.