Question

Subscribe 1. Matter of Shading One way to accurately render three-dimensional objects on a computer screen involves using the dot and cross products. In order to determine how to shade a piece of a surface we need to determine the angle at which rays from the light source hit the surface. To determine this angle, we compute the dot product of the light vector with the vector perpendicular to the surface at the particular point, called the normal vector light ray normal vector normal vector light ray This piece appears bright This piece appears dimmer If the light ray hits the surface straight on, that is, has an angle of 180° with the normal, then this piece of the surface will appear bright. On the other hand, if the light comes in on an angle, this piece of the surface will not bright appear as (a) Suppose the light source is placed directly above the xy-plane, so that the light rays come in parallel to the vector (0,0, 1). At what angle (to the normal) do the light rays hit a triangle bounded by the points (3,2,4), (2,5,3), and (1,2, 6)? Note that your answer will be an obtuse angle (greater than 90°) because the vertex of the angle is formed by the tails of the two vectors. (b) Repeat Problem 1, this time using the region bounded by (3,5,2), (3,3, 1), and (1,3, 1) (c) Suppose we are standing above the light source looking down on the xy-plane. Which of these two regions will appear brighter to us?

Answer 1

We have to find the equation of the plane formed by the 3 points given

Then we find the vector perpendicular to it. This is nothing but the normal vector.

Now dot product of the two vectors is used to find the angle formed by the light ray with the normal.

3 y- 2 z-4 0 -1 3 -1 -2 0 2

3x+2y+3z =25

To convert it in vector form

rn

3i 2j3k V32223 25 xiyzk) V322232

(xi+yizk) 3i+2j+3k V22 25 22

3i 2j 3k n= V22

which is the normal unit vector

Therefore. dot product of n and (0,0,1)

3i 23k(0i + 0j+1k)= cos0 22

3 cose 22

50%

2) Now for the second plane

3 y- 5 z- 2 0 0 -2 -1 -2 -2 -1

2y-4z=2

To convert it into vector form

02 4k V0222(4) 25 xiyzk) V0222 (4)2

$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).\left (\frac{2\hat{j}-4\hat{k}}{\sqrt{20}} \right )=\frac{25}{\sqrt{20}}$

$\hat{n}=\left (\frac{2\hat{j}-4\hat{k}}{\sqrt{20}} \right )$

dot product of the two unit vectors

$\left (\frac{2\hat{j}-4\hat{k}}{\sqrt{20}} \right ).(0\hat{i}+0\hat{j}+1\hat{k})=cos\theta$

$cos\theta =-\frac{4}{\sqrt{20}}$

$\theta =153.34^{\circ}$

This means this light ray is coming from below the normal vector

So, angle is 180-153.34 = 26.66^o

c)

We can see the second plane has lesser angle with the normal. So, the second plane has a brighter spot.