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$008 / 2 / 11(a)$ $A B C D$ is a parallelogram. $

$008 / 2 / 11(a)$ $A B C D$ is a parallelogram. $

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Question

008/2/11(a)008//2//11(a)
ABCDABCD is a parallelogram.
XX is the point on BCBC such that BX:XC2:1BX:XC-=2:1.
AB=2pvec(AB)=2p and AD=3qvec(AD)=3q.
Find, in terms of pp and qq.
(a) ACvec(AC),
Answer (a)AC(a) vec(AC)-=
III
(b) Axvec(Ax),
Answer (b) AXvec(AX)-=
[I]

Best Answer

I apologize, but the first question is incomplete. The number 008 is not written properly and there is no instruction or equation to solve.
For the second question:
(a) ACvec(AC)
Answer: ACvec(AC) is equal to AD+DCvec(AD)+ vec(DC). Since ABCDABCD is a parallelogram and AB=DCvec(AB)= vec(DC), then DC=2pvec(DC)=-2p. Thus, AC=AD+DC=3q+(2p)=3q2pvec(AC)= vec(AD)+ vec(DC)=3q+(-2p)=3q-2p.
(b) AXvec(AX)
Answer: Since BX:XC=2:1BX:XC=2:1, then BX=23BCvec(BX)=(2)/(3) vec(BC) and XC=13BCvec(XC)=(1)/(3) vec(BC). Since BC=CD+DB=AD+AB=2p3qvec(BC)= vec(CD)+ vec(DB)=- vec(AD)+ vec(AB)=2p-3q, then BX=23(2p3q)=43p2qvec(BX)=(2)/(3)(2p-3q)=(4)/(3)p-2q and XC=13(2p3q)=23pqvec(XC)=(1)/(3)(2p-3q)=(2)/(3)p-q. Thus, AX=AB+BX=2p+43p2q=103p2qvec(AX)= vec(AB)+ vec(BX)=2p+(4)/(3)p-2q=(10)/(3)p-2q.