Porisma

Worked solution

Distinguish four independent parameter cases

Relative to O, A and B have position vectors 2i+j+2k2\mathbf i+\mathbf j+2\mathbf k and 3i−2j+pk3\mathbf i-2\mathbf j+p\mathbf k, where p is a constant.

(a) Find p for which OA and OB are perpendicular. marks

(b) In the case p=6, find angle AOB to the nearest degree. marks

(c) In the case p=5, find the position vector of M on segment AB such that AM:MB=1:2. marks

(d) Find the values of p for which the length of AB is 11\sqrt{11} units. marks

[2008/JJC/Promo/11]

CJC · Review problems · Question 5

Answer

(a)−2; (b)40°; (c)OM=(7/3,0,3)OM=(7/3,0,3); (d)p=1 or 3.

(a) Perpendicular position vectors

OA⋅OB=6−2+2p=0OA\cdot OB=6-2+2p=0, so p=−2p=-2.

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(b) Angle at the origin when p=6

cos⁡∠AOB=(2,1,2)⋅(3,−2,6)949=16/21\cos\angle AOB=\frac{(2,1,2)\cdot(3,-2,6)}{\sqrt9\sqrt{49}}=16/21. Thus ∠AOB=cos⁡−1(16/21)=40∘\angle AOB=\cos^{-1}(16/21)=40^\circ to the nearest degree.

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(c) Weight the opposite endpoints when p=5

OM=(2OA+OB)/3=(2(2,1,2)+(3,−2,5))/3=(7/3,0,3)OM=(2OA+OB)/3=(2(2,1,2)+(3,-2,5))/3=(7/3,0,3).

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(d) Apply the stated displacement length

AB=OB−OA=(1,−3,p−2)AB=OB-OA=(1,-3,p-2). Thus 12+(−3)2+(p−2)2=11\sqrt{1^2+(-3)^2+(p-2)^2}=\sqrt{11}, giving 10+p2−4p+4=1110+p^2-4p+4=11, p2−4p+3=0p^2-4p+3=0, or (p−3)(p−1)=0(p-3)(p-1)=0. Hence p=1p=1 or 3.

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Collinear A,M,B in that order with AM:MB=1:2. M lies exactly one third of the way from A to B.
Collinear A,M,B in that order with AM:MB=1:2. M lies exactly one third of the way from A to B.
Collinear A,M,B in that order with AM:MB=1:2. M lies exactly one third of the way from A to B.

Common mistakes

A line-segment ratio weights the opposite endpoints. AB is a displacement, whereas OA and OB are position vectors.

Solution notes

Complete four-part working and the internal section diagram retained. The angle is rounded exactly as requested.

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