Porisma

Worked solution

Locate a perpendicular foot and reflect a whole line

(i) Find a vector equation of the line through the points AA and BB with position vectors 7i+8j+9k7\mathbf i+8\mathbf j+9\mathbf k and −i−8j+k-\mathbf i-8\mathbf j+\mathbf k respectively.

(ii) The perpendicular to this line from the point CC with position vector i+8j+3k\mathbf i+8\mathbf j+3\mathbf k meets the line at the point NN. Find the position vector NN and the ratio of AN:NBAN:NB.

(iii) Find a Cartesian equation of the line which is a reflection of the line ACAC in the line ABAB.

[2012/I/9]

CJC · Discussion · Question 7

Answer

AB:r=(7,8,9)+λ(1,2,1)AB:\mathbf r=(7,8,9)+\lambda(1,2,1); ON=(5,4,7)ON=(5,4,7); AN:NB=1:3AN:NB=1:3; reflected line x−7=(y−8)/(−4)=z−9x-7=(y-8)/(-4)=z-9.

(i) Equation of AB

AB=(−8,−16,−8)=−8(1,2,1)AB=(-8,-16,-8)=-8(1,2,1). Thus

r=(7,8,9)+λ(1,2,1),λ∈R.\mathbf r=(7,8,9)+\lambda(1,2,1),\quad\lambda\in\mathbb R.

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(ii) Find N using perpendicularity

N=(7+λ,8+2λ,9+λ)N=(7+\lambda,8+2\lambda,9+\lambda), so CN=(6+λ,2λ,6+λ)CN=(6+\lambda,2\lambda,6+\lambda). Its dot product with (1,2,1)(1,2,1) is 12+6λ=012+6\lambda=0. Thus λ=−2\lambda=-2 and

ON=(5,4,7).ON=(5,4,7).

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(ii) Determine the internal ratio

AN=(−2,−4,−2)AN=(-2,-4,-2) and NB=(−6,−12,−6)=3ANNB=(-6,-12,-6)=3 AN. Hence AN:NB=1:3AN:NB=1:3.

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(iii) Reflect C, keeping A fixed

Let C′C^{\prime} be the reflection of C in AB. Since N is its midpoint,

OC′=2ON−OC=(9,0,11).\begin{aligned}&OC^{\prime}\\ & =2 ON-OC\\ & =(9,0,11).\end{aligned}

Then AC′=(2,−8,2)=2(1,−4,1)AC^{\prime}=(2,-8,2)=2(1,-4,1). The reflected line is r=(7,8,9)+μ(1,−4,1)\mathbf r=(7,8,9)+\mu(1,-4,1), μ∈R\mu\in\mathbb R, giving

x−7=y−8−4=z−9.x-7=\frac{y-8}{-4}=z-9.

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Compare your sketch

Expand diagram

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A,N,B are collinear in that order with AN:NB=1:3. CN is perpendicular to AB and N is the midpoint of CC-prime. Segments AC and AC-prime are mirror images in AB. Planar schematic for the three-dimensional points.
A,N,B are collinear in that order with AN:NB=1:3. CN is perpendicular to AB and N is the midpoint of CC-prime. Segments AC and AC-prime are mirror images in AB. Planar schematic for the three-dimensional points.
A,N,B are collinear in that order with AN:NB=1:3. CN is perpendicular to AB and N is the midpoint of CC-prime. Segments AC and AC-prime are mirror images in AB. Planar schematic for the three-dimensional points.

Common mistakes

Reflecting AC requires a new line direction, not just the position vector of the reflected C.

Solution notes

All three source requests and the complete reflection construction retained. The independent diagram preserves perpendicularity, midpoint and internal 1:3 division.

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