Question #7a19c

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Answer 1 · 2017-03-11T09:13:03

drawn

Given #BC=2OA#

Parallelogram OADB comleted.

So #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)#