i. We know that #((p),(1),(1))# lies in the same 'plane' as #((4),(2),(p))#. One thing to notice is that the second number in #vec(OB)# is double that of #vec(OA)#, so #vec(OB)=2vec(OA)#
#((2p),(2),(2))=((4),(2),(p))#
#2p=4#
#p=2#
#2=p#
For the unit vector, we need a magnitude of 1, or #vec(OA)/abs(vec(OA))#. #abs(vec(OA))=sqrt(2^2+1+1)=sqrt6#
#hat(vec(OA))=1/sqrt6((2),(1),(1))=((2/sqrt6),(1/sqrt6),(1/sqrt6))=2/sqrt6i+1/sqrt6j+1/sqrt6k#
ii. #costheta=(veca.vecb)/(abs(veca)abs(vecb)#
#cos90=0#
So, #(veca.vecb)=0#
#vec(AB)=vec(OB)-vec(OA)=((4),(2),(p))-((p),(1),(1))=((4-p),(1),(p-1))#
#((p),(1),(1))*((4-p),(1),(p-1))=0#
#p(4-p)+1+p-1=0#
#p(4-p)-p=0#
#4p-p^2-p=0#
#3p-p^2=0#
#p(3-p)=0#
#p=0or3-p=0#
#p=0or3#
iii. #p=3#
#vec(OA)=((3),(1),(1))#
#vec(OB)=((4),(2),(3))#
A parallelogram has two sets of equal and opposite angles, so #C# must be located at #vec(OA)+vec(OB)# (I'll provide a diagram when possible).
#vec(OC)=vec(OA)+vec(OB)=((3),(1),(1))+((4),(2),(3))=((7),(3),(4))#
