Porisma

Worked solution

Find a perpendicular foot, shortest distance and reflected point

The point is P=(1,1,−1)P=(1,1,-1) and the line is r=(1,2,3)+λ(1,2,1)\mathbf r=(1,2,3)+\lambda(1,2,1), λ∈R\lambda\in\mathbb R.

(i) Find the coordinates of the foot of the perpendicular from P to the line.

(ii) Deduce the shortest distance from P to the line.

(iii) Find the position vector of the image of P in the line.

CJC · Lecture · Example 23

(i) Method 1: put F on the line and use a zero dot product

Write OF→=(1,2,3)+λ(1,2,1)\overrightarrow{OF}=(1,2,3)+\lambda(1,2,1). Then PF→=OF→−OP→=(λ,1+2λ,4+λ)\overrightarrow{PF}=\overrightarrow{OF}-\overrightarrow{OP}=(\lambda,1+2\lambda,4+\lambda). Since PF is perpendicular to the line, (λ,1+2λ,4+λ)⋅(1,2,1)=0(\lambda,1+2\lambda,4+\lambda)\cdot(1,2,1)=0. Thus λ+2+4λ+4+λ=0\lambda+2+4\lambda+4+\lambda=0, so λ=−1. Substitution gives OF→=(1,2,3)−(1,2,1)=(0,0,2)\overrightarrow{OF}=(1,2,3)-(1,2,1)=(0,0,2) and F=(0,0,2).

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(i) Method 2: project AP onto a unit direction

Choose A=(1,2,3), so AP→=(1,1,−1)−(1,2,3)=(0,−1,−4)\overrightarrow{AP}=(1,1,-1)-(1,2,3)=(0,-1,-4) and m^=(1,2,1)/6\widehat{\mathbf m}=(1,2,1)/\sqrt6. The projection vector is AF→=(AP→⋅m^)m^\overrightarrow{AF}=(\overrightarrow{AP}\cdot\widehat{\mathbf m})\widehat{\mathbf m}. Therefore OF→=OA→+(AP→⋅m^)m^=(1,2,3)+[(−2−4)/6](1,2,1)=(0,0,2)\overrightarrow{OF}=\overrightarrow{OA}+(\overrightarrow{AP}\cdot\widehat{\mathbf m})\widehat{\mathbf m}=(1,2,3)+[(-2-4)/6](1,2,1)=(0,0,2). The negative coefficient places F behind A in the positive m direction; the Teacher sketch is schematic and puts A on the other side of F.

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(ii) Take the magnitude of the perpendicular displacement

With λ=−1, PF→=(−1,1+2(−1),4−1)=(−1,−1,3)\overrightarrow{PF}=(-1,1+2(-1),4-1)=(-1,-1,3). Hence the shortest distance is PF=∣PF→∣=(−1)2+(−1)2+32=11PF=|\overrightarrow{PF}|=\sqrt{(-1)^2+(-1)^2+3^2}=\sqrt{11}.

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(iii) Use F as the midpoint of P and its image

Let the reflected point be P′. Since F is the midpoint of PP′, OF→=12(OP→+OP′→)\overrightarrow{OF}=\tfrac12(\overrightarrow{OP}+\overrightarrow{OP'}). Thus OP′→=2OF→−OP→=2(0,0,2)−(1,1,−1)=(−1,−1,5)\overrightarrow{OP'}=2\overrightarrow{OF}-\overrightarrow{OP}=2(0,0,2)-(1,1,-1)=(-1,-1,5). The two perpendicular segments PF and FP′ have equal length.

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Compare your sketch

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Three worked panels retain the Teacher constructions. P projects to F=(0,0,2); A=(1,2,3) is to the right of F in the direction m=(1,2,1), correcting the source schematic order. Reflecting across the line gives P prime=(-1,-1,5), with F the midpoint and equal perpendicular segments.
Three worked panels retain the Teacher constructions. P projects to F=(0,0,2); A=(1,2,3) is to the right of F in the direction m=(1,2,1), correcting the source schematic order. Reflecting across the line gives P prime=(-1,-1,5), with F the midpoint and equal perpendicular segments.
Three worked panels retain the Teacher constructions. P projects to F=(0,0,2); A=(1,2,3) is to the right of F in the direction m=(1,2,1), correcting the source schematic order. Reflecting across the line gives P prime=(-1,-1,5), with F the midpoint and equal perpendicular segments.

Common mistakes

The foot must both lie on the line and make PF perpendicular to its direction. A projection formula needs a unit direction (or division by m·m). Reflection uses 2 f−p, not f−p.

Solution notes

Both source methods and all three Teacher diagrams are incorporated. The projection drawing is corrected to the numerical order F,A; its original schematic placement remains documented in the native record.

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