The Student Room Group

Vectors

How is this true:

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

In http://www.examsolutions.net/maths-revision/core-maths/vectors/scalar-product/example-1.php at 3.30 he states it is a standard result can someone prove it please?
Reply 1
Original post by Substitution
How is this true:

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

In http://www.examsolutions.net/maths-revision/core-maths/vectors/scalar-product/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.
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
Reply 3
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.
Original post by Substitution
How is this true:

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

In http://www.examsolutions.net/maths-revision/core-maths/vectors/scalar-product/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
(edited 9 years ago)
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 :smile:

Quick Reply

Latest