3 points ABC on a plane, O as origins. OA=vec(a) , OB=vec(b) , OC=vec(c) . Point M inside Delta ABC. And Delta MAB: Delta MBC: Delta MCA=2:3:5. A straight line BM intersect side AC at N. Express OM in terms of vector a,b,c.

likovnihuj 2022-09-23 Answered
3 points ABC on a plane, O as origins. O A = a , O B = b , O C = c . Point M inside A B C. And M A B : M B C : M C A = 2 : 3 : 5
A straight line BM intersect side AC at N. Express OM in terms of vector a,b,c.
Can you give me some hint? I have been thinking, what i got is, B M : M N = 1 : 1. A B : B C = 2 : 3 A B = O B O A = b a C B = O B O C = b c
O M = O B + B M
B M = 1 2 B N Then, i have difficulity in expressing B N in terms of vector a,b,c.
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Answers (1)

Sarah Sutton
Answered 2022-09-24 Author has 4 answers
Note that from the given M A B : M B C : M C A = 2 : 3 : 5, the following relationship can be derived,
A N = 2 5 A C = 2 5 ( B C B A )
Then,
B N = A N A B = 2 5 ( B C B A ) + B A = 2 5 B C + 3 5 B A
where B C = c b and B A = a b . Thus,
O M = B M + O B = 1 2 B N + b = 3 10 a + 1 2 b + 1 5 c
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