Porisma

Worked solution

Find the perpendicular foot and reflect a point

Given three points A(0,2,7)A(0,2,7), B(5,−3,2)B(5,-3,2) and C(1,1,1)C(1,1,1), find the position vector of the point RR on ABAB such that CRCR is perpendicular to ABAB. Hence find the perpendicular distance from CC to ABAB and the position vector of the reflection of CC in ABAB.

CJC · Discussion · Question 4

Answer

OR=13(8,−2,13)OR=\frac 13(8,-2,13); distance 56/35\sqrt 6/3; reflected vector OC′=13(13,−7,23)OC^{\prime}=\frac 13(13,-7,23).

Parameterize R on AB

AB→=(5,−5,−5)=−5(−1,1,1)\overrightarrow{AB}=(5,-5,-5)=-5(-1,1,1). Thus r=(0,2,7)+λ(−1,1,1)\mathbf r=(0,2,7)+\lambda(-1,1,1), λ∈R\lambda\in\mathbb R. R has coordinates (−λ,2+λ,7+λ)(-\lambda,2+\lambda,7+\lambda) and CR=(−1−λ,1+λ,6+λ)CR=(-1-\lambda,1+\lambda,6+\lambda).

How did this part go?

Impose perpendicularity

Dot CR with (−1,1,1)(-1,1,1): 1+λ+1+λ+6+λ=01+\lambda+1+\lambda+6+\lambda=0. Hence λ=−8/3\lambda=-8/3 and

OR=13(8,−2,13).OR=\frac 13(8,-2,13).

How did this part go?

Calculate the distance

CR=53(1,−1,2)CR=\frac 53(1,-1,2), so

∣CR∣=531+1+4=563 units.\begin{aligned}&|CR|\\ & =\frac 53\sqrt{1+1+4}\\ & =\frac{5\sqrt 6}{3}\text{ units}.\end{aligned}

How did this part go?

Use the reflection midpoint

R is the midpoint of CC′, so

OC′=2OR−OC=13(13,−7,23).\begin{aligned}&OC^{\prime}\\ & =2 OR-OC\\ & =\frac 13(13,-7,23).\end{aligned}

How did this part go?

Compare your sketch

Expand diagram

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A(0,2,7),B(5,−3,2) lie on AB. R is the perpendicular foot from C(1,1,1) and midpoint of CC-prime. The figure depicts the plane of A,B,C and the reflection, not x/y axes for the coordinate triples.
A(0,2,7),B(5,−3,2) lie on AB. R is the perpendicular foot from C(1,1,1) and midpoint of CC-prime. The figure depicts the plane of A,B,C and the reflection, not x/y axes for the coordinate triples.
A(0,2,7),B(5,−3,2) lie on AB. R is the perpendicular foot from C(1,1,1) and midpoint of CC-prime. The figure depicts the plane of A,B,C and the reflection, not x/y axes for the coordinate triples.

Common mistakes

Use C′=2 R−C. The opening sentence contains three requests even though they are unnumbered.

Solution notes

All source requests, working and reflection geometry retained. The independent diagram uses spatial coordinate labels in their common plane.

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