Porisma

Worked solution

Project a pyramid shadow under two different lighting models

A right opaque pyramid has square base ABCD at ground level, centre O and vertex V. Perpendicular unit vectors i,j,k\mathbf i,\mathbf j,\mathbf k point along AB, AD, OV, respectively. AB has length 8 and OV has length 2h. A point light source is at P(20,−4,0)P(20,-4,0). A screen of height 35 has its base on the ground and lies in r⋅(1,0,0)=α\mathbf r\cdot(1,0,0)=\alpha, where α<−4\alpha<-4, as shown.

(i) Find a vector equation for the line depicting the light ray from P to V, in terms of h. marks

(ii) Find an inequality between α,h\alpha,h so the pyramid’s shadow does not exceed the screen height. marks

The point light is now replaced by parallel light rays perpendicular to the screen; also h=10h=10.

(iii) Find the exact length of the shadow cast by edge VB on the screen. marks

A mirror is placed on plane VBC.

(iv) Find a vector equation of plane VBC and hence the mirror’s inclination to the ground. marks

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Given setup: square ABCD on the ground centred at O, vertex V vertically above O, i along AB, j along AD, k along OV. P(20,-4,0) is beyond B on the AB ground line. The screen stands on x=alpha less than -4 and has height 35; base AB=8 and OV=2h.
Given setup: square ABCD on the ground centred at O, vertex V vertically above O, i along AB, j along AD, k along OV. P(20,-4,0) is beyond B on the AB ground line. The screen stands on x=alpha less than -4 and has height 35; base AB=8 and OV=2h.
Given setup: square ABCD on the ground centred at O, vertex V vertically above O, i along AB, j along AD, k along OV. P(20,-4,0) is beyond B on the AB ground line. The screen stands on x=alpha less than -4 and has height 35; base AB=8 and OV=2h.

[2017/RI/I/11]

CJC · Discussion · Question 10

Answer

(i) r=(20,−4,0)+λ(−10,2,h)\mathbf r=(20,-4,0)+\lambda(-10,2,h); (ii) h≤350/(20−α)h\le350/(20-\alpha); (iii) 4264\sqrt{26}; (iv) r⋅(5,0,1)=20\mathbf r\cdot(5,0,1)=20, inclination 78.7∘78.7^\circ.

(i) Join the point source to the vertex

With origin O, P=(20,−4,0)P=(20,-4,0) and V=(0,0,2h)V=(0,0,2h). Thus PV→=(−20,4,2h)=2(−10,2,h)\overrightarrow{PV}=(-20,4,2h)=2(-10,2,h) and the line is r=(20,−4,0)+λ(−10,2,h)\mathbf r=(20,-4,0)+\lambda(-10,2,h), λ∈R\lambda\in\mathbb R. The physical forward ray uses λ≥0\lambda\ge0, and passes V at λ=2\lambda=2.

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(ii) Intersect the ray with the screen

The screen has x=αx=\alpha, so 20−10λ=α20-10\lambda=\alpha and λ=(20−α)/10\lambda=(20-\alpha)/10. The vertex shadow height is λh=(20−α)h/10\lambda h=(20-\alpha)h/10. Since α<−4\alpha<-4, 20−α>020-\alpha>0, so the height bound is h≤350/(20−α)h\le350/(20-\alpha) (with the physical pyramid height h>0h>0). This uses point-source rays; it is not the later parallel projection.

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(iii) Switch to parallel orthogonal projection

Now B=(4,−4,0)B=(4,-4,0), V=(0,0,20)V=(0,0,20) and BV→=(−4,4,20)\overrightarrow{BV}=(-4,4,20). Rays parallel to the unit screen normal (1,0,0)(1,0,0) remove the x-component: the shadow displacement is (0,4,20)(0,4,20). Equivalently, the source uses ∣BV→×(1,0,0)∣=∣(0,20,−4)∣=416=426|\overrightarrow{BV}\times(1,0,0)|=|(0,20,-4)|=\sqrt{416}=4\sqrt{26}. Both describe the same projected length.

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(iv) Construct the mirror plane and compare normals

C=(4,4,0)C=(4,4,0) gives CV→=(−4,−4,20)=−4(1,1,−5)\overrightarrow{CV}=(-4,-4,20)=-4(1,1,-5), and BV→=4(−1,1,5)\overrightarrow{BV}=4(-1,1,5). Crossing (1,1,−5)(1,1,-5) and (−1,1,5)(-1,1,5) gives (10,0,2)=2(5,0,1)(10,0,2)=2(5,0,1). Therefore the plane is r⋅(5,0,1)=V⋅(5,0,1)=20\mathbf r\cdot(5,0,1)=V\cdot(5,0,1)=20, or 5x+z=205x+z=20. The ground normal is (0,0,1)(0,0,1), so the inclination is cos⁡−1(1/26)≈78.7∘\cos^{-1}(1/\sqrt{26})\approx78.7^\circ (1 d.p.).

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Common mistakes

Part (iii) changes the light source; do not reuse the point-source enlargement factor.

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