# VectorsWatch

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Thread starter 5 years ago
#1
How is this true:

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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5 years ago
#2
(Original post by Substitution)
How is this true:

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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Thread starter 5 years ago
#3
(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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5 years ago
#4
(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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5 years ago
#5
(Original post by Substitution)
How is this true:

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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Thread starter 5 years ago
#6
(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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