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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. |
Answer» 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`. |
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