i. Prove that the points `veca - 2vecb + 3 vecc, 2 veca + 3vecb- 4 vecc and -7 vecb + 10 vecc` are collinear, where `veca, vec b and vecc` are non-coplanar.
ii. Prove that the points `A(1, 2, 3), B(3,4, 7) and C(-3, -2, -5)` are collinear. Find the ratio in which point C divides AB.

Explanation

Let the given points be A, B and C. Therefore,
`" " vec(AB)` = P.V. of B- P.V. of A
`" " = ( 2veca + 3 vecb - 4vecc) - (vec a - 2 vecb + 3vecc)`
`" " = veca + 5 vecb - 7 vecc `
`" " vec(AC) `= P.V of C - P.V of A
`" "=(-7vecb + 10 vec c) - (veca - 2vecb + 3 vecc)`
`" " = - veca - 5 vecb + 7vecc = - vec(AB)`
Since `vec(AC) =- vec(AB)`, it follows that the points A, B and C are collinear.
ii. Let C divide AB in the ratio `k:1`, then `C(-3, -2, -5) -= ((3k+1)/(k+1), (4k+2)/(k+1), (7k+3)/(k+1))`
`rArr " " (3k+1)/(k+1) = -3, (4k+2)/(k+1) = -2 and (7k+3)/(k+1) =-5`
`rArr " " k = -(2)/(3) ` from all relations
Hence, C divides AB externally in the ratio `2 : 3`.