# Using some simple trigonometry we find the line from

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Using some simple trigonometry, we find the line from the source to ( t, 0) has length l 1 = h 2 + t 2 and the line from there to the viewer has length 1

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l 2 = y 2 + ( x t ) 2 . The total time light travels is thus l 1 v 1 + l 2 v 2 . Minimizing over t gives desired result when we note the two desired sines are sin θ 1 = h h 2 + t 2 and sin θ 2 = y ( y 2 +( x t ) 2 . 6.19 Shading requires that when we transform normals and points, we maintain the angle between them or equivalently have the dot product p · v = p · v when p = Mp and n = Mp . If M T M is an identity matrix angles are preserved. Such a matrix ( M 1 = M T ) is called orthogonal. Rotations and translations are orthogonal but scaling and shear are not. 6.21 Probably the easiest approach to this problem is to rotate the given plane to plane z = 0 and rotate the light source and objects in the same way. Now we have the same problem we have solved and can rotate everything back at the end. 6.23 A global rendering approach would generate all shadows correctly. In a global renderer, as each point is shaded, a calculation is done to see which light sources shine on it. The projection approach assumes that we can project each polygon onto all other polygons. If the shadow of a given polygon projects onto multiple polygons, we could not compute these shadow polygons very easily. In addition, we have not accounted for the different shades we might see if there were intersecting shadows from multiple light sources. 2
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