Question #7a19c

Question

1833 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-11T09:13:03

drawn

Given $B C = 2 O A$ $BC=2OA$

Parallelogram OADB comleted.

So $- - \to B D = - - \to D C = p$ $vec(BD)=vec(DC)=p$

By triangle law of vector addition for $Delta ADC$

$vec(AC)=vec(AD)+vec(DC)=vec(OB)+vec(OA)=q+p$

Again by triangle law of vector addition for $Delta OAM$

$vec(OM)=vec(OA)+vec(AM)=vec(OA)+1/2vec(AC)=p+1/2(q+p)$

$=1/2(q+3p)$

Answer 2 · 2017-03-11T09:14:52

a) $" "vec(AC)=vecp+vecq$

b) $" "vec(OM)=1/2(3vecp+vecq)$

a)

$vec(AC)=vec(AO)+vec(OB)+vec(BC)----(1)$

now we are told that $BCuarr uarr$ and $2xx$ OA#

$:.vec(BC)=2vecp$

$(1)" becomes "$

$vec(AC)=-vecp+vecq+2vecp$

$vec(AC)=vecp+vecq$

b)

now fro the diagram $M$ is the midpoint of AC

$:.vec(AM)=1/2vec(AC)=1/2(vecp+vecq)$

$vec(OM)=vec(OA)+vec(AM)$

$vec(OM)=vecp+1/2(vecp+vecq)$

$vec(OM)=1/2(3vecp+vecq)$