Porisma

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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

Set the screen coordinate x to alpha, then distinguish diverging point-source rays from the later parallel rays.

Read the worked solution