Porisma

Worked solution

Use an ordered parallelogram and a norm constraint

Referred to O, A,B have position vectors a=i−j+k\mathbf a=\mathbf i-\mathbf j+\mathbf k, b=2i+3k\mathbf b=2\mathbf i+3\mathbf k. C has position vector c=λa+μb\mathbf c=\lambda\mathbf a+\mu\mathbf b, where λ,μ are constants.

(i) Find the exact area of triangle OAB. marks

(ii) Given that OABC forms a parallelogram, write down λ and μ. marks

(iii) Given instead μ=2 and OC=59OC=\sqrt{59}, find the possible coordinates of C in exact form. marks

[2016/MJC/Promo/4]

CJC · Review problems · Question 8

Answer

Area=14/2\sqrt{14}/2; (ii)λ=−1,μ=1; (iii)C=(13/3,−1/3,19/3)C=(13/3,-1/3,19/3) or (−3,7,−1)(-3,7,-1).

(i) Compute a triangular area

a×b=(−3,−1,2)\mathbf a\times\mathbf b=(-3,-1,2), so area⁡(OAB)=129+1+4=14/2\operatorname{area}(OAB)=\frac12\sqrt{9+1+4}=\sqrt{14}/2.

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(ii) Follow the order O–A–B–C

OC=AB=b−a=λa+μbOC=AB=\mathbf b-\mathbf a=\lambda\mathbf a+\mu\mathbf b. Therefore λ=−1 and μ=1.

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(iii) Solve the independent norm condition

With μ=2, c=(λ+4,−λ,λ+6)\mathbf c=(\lambda+4,-\lambda,\lambda+6). The length constraint gives (λ+4)2+λ2+(λ+6)2=59(\lambda+4)^2+\lambda^2+(\lambda+6)^2=59, or 3λ2+20λ−7=(3λ−1)(λ+7)=03\lambda^2+20\lambda-7=(3\lambda-1)(\lambda+7)=0. Thus λ=1/3 or −7 and C=(13/3,−1/3,19/3)C=(13/3,-1/3,19/3) or (−3,7,−1)(-3,7,-1).

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Common mistakes

Part (iii) says instead; discard the constants found in(ii).

Solution notes

Every printed ask and all exact algebra retained; this question has no source figure.

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