Porisma

Worked solution

Reflect an intersecting line by reflecting one point

Line l1l_1 passes through A=(−1,−2,1), B=(0,1,5). Line l2l_2 has equation x−1=(y−2)/2=z−3x-1=(y-2)/2=z-3. They intersect at A. Find:

(i) vector equations of l1,l2l_1,l_2; marks

(ii) their acute angle; marks

(iii) the position vector of the perpendicular foot F from B onto l2l_2; marks

(iv) the equation of the line l3l_3, the mirror image of l1l_1 in l2l_2. marks

[2012/CJC/Promo/10(a)]

CJC · Review problems · Question 7

Answer

l1:r=(0,1,5)+λ(1,3,4)l_1:r=(0,1,5)+\lambda(1,3,4); l2:r=(1,2,3)+μ(1,2,1)l_2:r=(1,2,3)+\mu(1,2,1); θ=28.3°; OF=(5,10,17)/6OF=(5,10,17)/6; l3:r=(−1,−2,1)+α(8,13,−1)l_3:r=(-1,-2,1)+\alpha(8,13,-1).

(i) Write independent parameterisations

AB=(1,3,4)AB=(1,3,4). Thus l1:r=(0,1,5)+λ(1,3,4)l_1:\mathbf r=(0,1,5)+\lambda(1,3,4) and l2:r=(1,2,3)+μ(1,2,1)l_2:\mathbf r=(1,2,3)+\mu(1,2,1), with λ,μ real. A lies on l2 at μ=−2.

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(ii) Use the acute angle formula

cos⁡θ=∣(1,3,4)⋅(1,2,1)∣/(266)=11/156\cos\theta=|(1,3,4)\cdot(1,2,1)|/(\sqrt{26}\sqrt6)=11/\sqrt{156}. Thus θ=28.272…°=28.3°.

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(iii) Find a foot by a perpendicular displacement

OF=(1,2,3)+μ(1,2,1)OF=(1,2,3)+\mu(1,2,1) and BF=(1,1,−2)+μ(1,2,1)BF=(1,1,-2)+\mu(1,2,1). Setting BF⋅(1,2,1)=0BF\cdot(1,2,1)=0 gives 1+μ+2+4μ+μ−2=01+\mu+2+4\mu+\mu-2=0, so μ=−1/6 and OF=(5,10,17)/6OF=(5,10,17)/6.

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(iv) Reflect B and keep A fixed

If B′ is the image of B, OB′=2OF−OB=(5,10,17)/3−(0,1,5)=(5,7,2)/3OB′=2OF-OB=(5,10,17)/3-(0,1,5)=(5,7,2)/3. Hence AB′=(5,7,2)/3−(−1,−2,1)=(8,13,−1)/3AB′=(5,7,2)/3-(-1,-2,1)=(8,13,-1)/3. Since A lies on the mirror line, it is fixed, so l3:r=(−1,−2,1)+α(8,13,−1)l_3:\mathbf r=(-1,-2,1)+\alpha(8,13,-1), α∈R\alpha\in\mathbb R.

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Common mistakes

The source labels its last solution (v), but the corresponding printed ask is (iv).

Solution notes

All 4 asks and complete foot/reflection algebra retained. The source final working label(v) is a numbering typo, retained in native text and mapped to ask(iv). No figure.

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