Substitution
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How is this true:

\vec{AB} = \vec{OA} - \vec{OB}

In http://www.examsolutions.net/maths-r.../example-1.php at 3.30 he states it is a standard result can someone prove it please?
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davros
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(Original post by Substitution)
How is this true:

\vec{AB} = \vec{OA} - \vec{OB}

In http://www.examsolutions.net/maths-r.../example-1.php at 3.30 he states it is a standard result can someone prove it please?
Are you sure it isn't AB = OB - OA? That would be a standard result.
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Substitution
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(Original post by davros)
Are you sure it isn't AB = OB - OA? That would be a standard result.
It is yes, can you explain why or is it just something to know? Thanks
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james22
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(Original post by Substitution)
It is yes, can you explain why or is it just something to know? Thanks
Trying writing it out in terms of coordinates, it becomes much clearer then.
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physics4ever
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(Original post by Substitution)
How is this true:

\vec{AB} = \vec{OA} - \vec{OB}

In http://www.examsolutions.net/maths-r.../example-1.php at 3.30 he states it is a standard result can someone prove it please?

just got this from google images, going from A to B (AB) is the same as going from O to C (OC) so AB=OC also you can get from O to C by taking the route from O to B to C
(OB + BC), but BC = -OA so OC= OB - OA =AB

you can see just by looking at the parrallelogram that
AB=OC=OB+BC=OB-OA
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Substitution
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(Original post by physics4ever)

just got this from google images, going from A to B (AB) is the same as going from O to C (OC) so AB=OC also you can get from O to C by taking the route from O to B to C
(OB + BC), but BC = -OA so OC= OB - OA =AB

you can see just by looking at the parrallelogram that
AB=OC=OB+BC=OB-OA
Really helpful thanks
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