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June 3rd, 2017, 01:10 PM   #1
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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!
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June 3rd, 2017, 01:16 PM   #2
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what are you trying to do?
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June 3rd, 2017, 01:33 PM   #3
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Quote:
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what are you trying to do?
–ěh, sorry I forgot to say that. I need to express vector AS in terms of vectors a and b
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June 3rd, 2017, 03:28 PM   #4
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I got -(3/13)a + (12/13)b, but I haven't checked your attempt yet.
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June 4th, 2017, 02:08 AM   #5
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Quote:
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I got -(3/13)a + (12/13)b, but I haven't checked your attempt yet.
That is exactly the answer in the book. Can you please check my answer or write yours here. It will mean a world to me
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June 4th, 2017, 03:06 AM   #6
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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.
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June 4th, 2017, 05:06 AM   #7
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Quote:
Originally Posted by skipjack View Post
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.
Okay, now I understand it. Thank you veryyy much
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