A wardrobe model from lines and planes

A home interior designer is looking to design a pull-down wardrobe lift. To test his concept, the designer built a small-scale model of it, in which points (x,y,z)(x,y,z) are defined relative to a corner of the model at O(0,0,0)O(0,0,0). The rods used in the model can be modelled by equations of straight lines.

The main rod can be modelled by the line l1l_1 with equation r=(5,−5,8)+λ(1,1,0)\mathbf r=(5,-5,8)+\lambda(1,1,0), λ∈R\lambda\in\mathbb R. Another rod, line l2l_2, passes through the points A(5,−5,8)A(5,-5,8) and B(8,−4,3)B(8,-4,3).

(a) A supporting wire connects BB to a point FF on l1l_1 such that BFBF is perpendicular to l1l_1. Find the position vector of FF. marks

(b) A retractable rod, line lRl_R, is placed such that it can be modelled by reflecting l2l_2 about l1l_1. Find a vector equation of lRl_R. marks

The panels of the small-scale model can be modelled by equations of planes. The main panel, plane p1p_1, contains the main rod l1l_1 and an attachment point C(0,0,6)C(0,0,6). A side panel, plane p2p_2, can be represented by the equation p2:−17x−37y+4z=24p_2:-17x-37y+4z=24.

(c) Show that p1p_1 can be represented by the cartesian equation −x+y+5z=30-x+y+5z=30. marks

(d) A structural rod, line lSl_S, needs to be placed at the intersection between p1,p2p_1,p_2. Find a vector equation of lSl_S. marks

(e) A decorative plank, plane p3p_3, has equation p3:−3x+αy+15z=βp_3:-3x+\alpha y+15z=\beta, where α,β∈R\alpha,\beta\in\mathbb R. If the distance between p1,p3p_1,p_3 is exactly 333\sqrt3 units, find the possible values of α,β\alpha,\beta. marks

[2023/TMJC/Prelim/I/9]

Total marks: 14

Solution

(a)

Foot of a perpendicular

l1:r=(5,−5,8)+λ(1,1,0)l_1:\mathbf r=(5,-5,8)+\lambda(1,1,0) and B(8,−4,3)B(8,-4,3). Find the foot FF of the perpendicular from BB to l1l_1.

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Hint

  1. F=(5,−5,8)+λ(1,1,0)F=(5,-5,8)+\lambda(1,1,0) with BF→⋅(1,1,0)=0\overrightarrow{BF}\cdot(1,1,0)=0.

Answer · (a)

F=(7,−3,8)F=(7,-3,8).

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The wire BF

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(b)

Reflecting a line

Find a vector equation of lRl_R, the reflection of l2l_2 (through AA and BB) in l1l_1.

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Hint

  1. FF is the midpoint of BB and its image B′B'.

  2. lRl_R passes through AA and B′B'.

Answer · (b)

lR:r=(5,−5,8)+t(1,3,5)l_R:r=(5,-5,8)+t(1,3,5).

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Move B and reflect the rod about l₁

A = (5, −5, 8) lies on the fixed mirror line. Move B to compare AB with AB′; the perpendicular foot F remains the midpoint of BB′. Reset restores the original rods.

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(c)

Plane through a line and a point

p1p_1 contains l1l_1 and C(0,0,6)C(0,0,6). Show p1:−x+y+5z=30p_1:-x+y+5z=30.

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Hint

  1. Normal =CA→×(1,1,0)=\overrightarrow{CA}\times(1,1,0).

Answer · (c)

p1:−x+y+5z=30p_1:-x+y+5z=30.

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The plane p₁

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(d)

Intersection of planes

p2:−17x−37y+4z=24p_2:-17x-37y+4z=24. Find a vector equation of lS=p1∩p2l_S=p_1\cap p_2.

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Hint

  1. Direction =n1×n2=\mathbf n_1\times\mathbf n_2.

  2. Find a common point.

Answer · (d)

lS:r=(−21,9,0)+t(7,−3,2)l_S:r=(-21,9,0)+t(7,-3,2).

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The structural rod l_S

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(e)

Parallel planes at a given distance

p3:−3x+αy+15z=βp_3:-3x+\alpha y+15z=\beta is 333\sqrt3 from p1p_1. Find α\alpha, β\beta.

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Hint

  1. Parallel to p1p_1: (−3,α,15)=3(−1,1,5)(-3,\alpha,15)=3(-1,1,5).

  2. Distance between −x+y+5z=30-x+y+5z=30 and −x+y+5z=β3-x+y+5z=\frac\beta3.

Answer · (e)

α=3\alpha=3; β=9\beta=9 or 171171.

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Two planes at distance 3√3

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