Question

Points l,m,n divides sides bc,ca,ab of triangle abc in the ratio s 1:4,3:2,3:7 respectively. prove that al + bm +cn is a vector parallel to ck where k divides ab in ratio 1:3

Answer 1

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Answer 2

Ratio of the ΔABC = 1:4, 3:2 , 3:7 (Given)

Let the position of vertices A B C of ΔABC be = a b and c

P.v pf L = c + 4b/5

P.v pf M = 3a + 2c/5

P.v pf N = 3b + 7a /10

AL + BM + CN = (c + 4b/5 - a) + (3a + 2c/5 -b) + (3b + 7a /10-c)

= 1/10 ( 3a + b - 4c)

Therefore, PV of K = 3a + b/4

CK = 3a +b/4 - c = 3a + b - 4c/4  

= 10/4 ( 3a + b - 4c/10) = 5/2 AL + BM + CN

= AL + BM + CN = 2/5CK

Thus, AL + BM + CN is parallel to vector CK.