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

likovnihuj

Answered question

2022-09-23

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.

Answer & Explanation

Sarah Sutton

Sarah Sutton

Beginner2022-09-24Added 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

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?