Porisma

Worked solution

Model cable intersections and minimise drilling distance

Electrical engineers are installing electricity cables on a building site. Points (x,y,z)(x,y,z) are defined relative to a main switching site at (0,0,0)(0,0,0), where units are metres. Cables are laid in straight lines and the widths of cables can be neglected.

An existing cable CC starts at the main switching site and goes in the direction (3,1,−2)(3,1,-2).

A new cable is installed which passes through points P(1,2,−1)P(1,2,-1) and Q(5,7,a)Q(5,7,a).

(i) Find the value of aa for which CC and the new cable will meet.

To ensure that the cables do not meet, the engineers use a=−3a=-3. The engineers wish to connect each of the points PP and QQ to a point RR on CC.

(ii) The engineers wish to reduce the length of cable required and believe in order to do this that angle PRQPRQ should be 90∘90^\circ. Show that this is not possible.

(iii) The engineers discover that the ground between PP and RR is difficult to drill through and now decide to make the length of PRPR as small as possible. Find the coordinates of RR in this case and the exact minimum length.

[2017/I/10]

CJC · Discussion · Question 10

Answer

(i) a=−22/5a=-22/5 for the full supporting lines (the source convention); no intersection on the forward ray. (ii) RP⋅RQ>0RP\cdot RQ>0 for every real λ\lambda. (iii) OR=(3/2,1/2,−1)OR=(3/2,1/2,-1) and minimum PR=10/2PR=\sqrt{10}/2 m.

(i) Intersect the supporting lines

Following the printed solution, write C:r=λ(3,1,−2)C:\mathbf r=\lambda(3,1,-2) and the new cable r=(1,2,−1)+μ(4,5,a+1)\mathbf r=(1,2,-1)+\mu(4,5,a+1), with real parameters. The equations

3λ−4μ=1,λ−5μ=2,−2λ=−1+(a+1)μ3\lambda-4\mu=1,\quad\lambda-5\mu=2,\quad-2\lambda=-1+(a+1)\mu

give μ=−5/11\mu=-5/11, λ=−3/11\lambda=-3/11 and a=−22/5a=-22/5. This is the source answer for full lines. If “starts at O and runs in the direction” means only the forward ray λ≥0\lambda\ge 0, there is no intersection for any a: the first two coordinates already force negative λ\lambda. Likewise the computed point is outside the finite segment PQ.

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(ii) Test the right angle

Now a=−3a=-3 and R=(3λ,λ,−2λ)R=(3\lambda,\lambda,-2\lambda). Thus RP=(1−3λ,2−λ,2λ−1)RP=(1-3\lambda,2-\lambda,2\lambda-1) and RQ=(5−3λ,7−λ,2λ−3)RQ=(5-3\lambda,7-\lambda,2\lambda-3). Their dot product is

RP⋅RQ=14λ2−35λ+22=14[(λ−54)2+1112]>0.\begin{aligned}&RP\cdot RQ\\ & =14\lambda^2-35\lambda+22\\ & =14\left[\left(\lambda-\frac 54\right)^2+\frac 1{112}\right]>0.\end{aligned}

It cannot be zero, so the angle cannot be 90∘90^\circ. Alternatively its discriminant is (−35)2−4(14)(22)=−7<0(-35)^2-4(14)(22)=-7<0, and its leading coefficient is positive. Its vertex is (5/4,1/8)(5/4,1/8); the source blue note incorrectly prints height 1/281/28.

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(iii) Form the squared distance

PR2=(3λ−1)2+(λ−2)2+(1−2λ)2=14λ2−14λ+6.\begin{aligned}&PR^2\\ & =(3\lambda-1)^2+(\lambda-2)^2\\ &{}+(1-2\lambda)^2\\ & =14\lambda^2-14\lambda+6.\end{aligned}

Minimising this nonnegative quantity is equivalent to minimising PR.

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(iii) Method 1: complete the square

PR2=14(λ−12)2+52.PR^2=14\left(\lambda-\frac 12\right)^2+\frac 52.

The minimum occurs at λ=1/2\lambda=1/2.

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(iii) Method 2 A: differentiate PR

d(PR)dλ=28λ−14214λ2−14λ+6.\begin{aligned}&\frac{d(PR)}{d\lambda}\\ & =\frac{28\lambda-14}{2\sqrt{14\lambda^2-14\lambda+6}}.\end{aligned}

The denominator is positive, and the source sign chart is:

λ\lambda<1/2<1/21/21/2>1/2>1/2
d(PR)/dλd(PR)/d\lambda−-00++
PRPRdecreasingminimumincreasing

Thus PR is minimised at λ=1/2\lambda=1/2 without needing the lengthy second derivative.

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(iii) Method 2 B: differentiate the square

d(PR2)dλ=28λ−14,d2(PR2)dλ2=28>0.\frac{d(PR^2)}{d\lambda}=28\lambda-14,\qquad\frac{d^2(PR^2)}{d\lambda^2}=28>0.

λ\lambda<1/2<1/21/21/2>1/2>1/2
d(PR2)/dλd(PR^2)/d\lambda−-00++
PR2PR^2decreasingminimumincreasing

Both the first-derivative sign change and the positive second derivative identify the same minimum.

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(iii) Method 3: perpendicular foot

The shortest connector to a line is perpendicular to its direction. With PR=(3λ−1,λ−2,1−2λ)PR=(3\lambda-1,\lambda-2,1-2\lambda),

PR⋅(3,1,−2)=3(3λ−1)+(λ−2)−2(1−2λ)=14λ−7=0.\begin{aligned}&PR\cdot(3,1,-2)\\ & =3(3\lambda-1)+(\lambda-2)-2(1-2\lambda)\\ & =14\lambda-7\\ & =0.\end{aligned}

Thus λ=1/2\lambda=1/2. This is allowed even if C is restricted to its forward ray.

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(iii) Evaluate the minimum

All methods give

OR=(32,12,−1),PR=14+94=102 metres.OR=\left(\frac 32,\frac 12,-1\right),\qquad PR=\sqrt{\frac 14+\frac 94}=\frac{\sqrt{10}}2\text{ metres}.

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Compare your sketch

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A planar schematic of the cable line C with direction (3,1,-2). R=(3/2,1/2,-1) is the perpendicular foot from P=(1,2,-1), and PR=sqrt(10)/2 metres is the minimum drilling distance. This is not an x/y coordinate plot.
A planar schematic of the cable line C with direction (3,1,-2). R=(3/2,1/2,-1) is the perpendicular foot from P=(1,2,-1), and PR=sqrt(10)/2 metres is the minimum drilling distance. This is not an x/y coordinate plot.
A planar schematic of the cable line C with direction (3,1,-2). R=(3/2,1/2,-1) is the perpendicular foot from P=(1,2,-1), and PR=sqrt(10)/2 metres is the minimum drilling distance. This is not an x/y coordinate plot.

Common mistakes

The full-line convention in part (i) matters: its computed parameter is negative. Keep that source modelling ambiguity explicit.

Solution notes

All printed asks, complete-square and discriminant arguments, three minimisation methods with both derivative variants, two sign charts and the perpendicular-foot sketch retained. The original vertex-height slip 1/28 is qualified; the correct value is 1/8. Source cable wording interpreted conditionally as full supporting lines, not silently changed to a forward ray.

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