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June 3rd, 2017, 12:10 PM  #1 
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  I need a little help with vectors.
Okay so, P is bisecting ED and R is bisecting FE. This is my attempt but the answer in the books says it's not true. Can somebody tell me what I do wrong? Thank you!

June 3rd, 2017, 12:16 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,412 Thanks: 716 
what are you trying to do?

June 3rd, 2017, 12:33 PM  #3 
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  
June 3rd, 2017, 02:28 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
I got (3/13)a + (12/13)b, but I haven't checked your attempt yet.

June 4th, 2017, 01:08 AM  #5 
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  
June 4th, 2017, 02:06 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
I assume the hexagon is regular. With A as origin, vector AS is a fraction, m, of vector AP, i.e. m((1/2)a + 2b). Similarly, vector RS is a fraction, n, of vector RB, i.e. n(2a  (3/2)b), so vector AS = AR + RS = a + (3/2)b + n(2a  (3/2)b). As these two expressions for vector AS are equivalent, m/2 = 2n  1 and 2m = 3/2  (3/2)n, so m = 6/13 and n = 5/13. Substituting these values into either expression for vector AS gives (3/13)a + (12/13)b. Your method was along much the same lines, but you found vector BR incorrectly. 
June 4th, 2017, 04:06 AM  #7  
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  Quote:
 

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