Exercise Solutions for Introduction to 3D Game Programming with DirectX 10

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1 Exercise Solutions for Introduction to 3D Game Programming with DirectX 10 Frank Luna, September 6, 009 Solutions to Part I Chapter 1 1. Let u = 1, and v = 3, 4. Perform the following computations and draw the vectors relative to a D coordinate system. a) u + v b) u v c) u + 1 v d) u + v a) 1, + 3, 4 = 1 + 3, + 4 = 4, b) 1, 3, 4 = 1, + 3, 4 = 1 3, + 4 =, 6 c) 1, + 1 3, 4 =, 4 + 3, = 7, d) 1, + 3, 4 =, 4 + 3, 4 = 1, 8. Let u = 1, 3, and v = 3, 4, 1. Perform the following computations. a) u + v b) u v c) 3u + v d) u + v a) 1, 3, + 3, 4, 1 = 1 + 3, 3 + 4, + 1 =, 1, 3 b) 1, 3, 3, 4, 1 = 1, 3, + 3, 4, 1 = 4, 7, 1 c) 3 1, 3, + 3, 4, 1 = 3, 9, 6 + 6, 8, = 3, 1, 8 d) 1, 3, + 3, 4, 1 =, 6, 4 + 3, 4,1 = 5, 10, 3

2 3. This exercise shows that vector algebra shares many of the nice properties of real numbers (this is not an exhaustive list). Assume u = u x, u y, u z, v = v x, v y, v z, and w = w x, w y, w z. Also assume that c and k are scalars. Prove the following vector properties. a) b) c) d) a) u + v = v + u (Commutative Property of Addition) b) u + v + w = u + v + w (Associative Property of Addition) c) ck u = c ku (Associative Property of Scalar Multiplication) d) k u + v = ku + kv (Distributive Property 1) e) u k + c = ku + cu (Distributive Property ) u + v = u x, u y, u z + v x, v y, v z = u x + v x, u y + v y, u z + v z = v x + u x, v y + u y, v z + u z = v x, v y, v z + u x, u y, u z = v + u u + v + w = u x, u y, u z + v x, v y, v z + w x, w y, w z = u x, u y, u z + v x + w x, v y + w y, v z + w z = u x + v x + w x, u y + v y + w y, u z + v z + w z = u x + v x + w x, u y + v y + w y, u z + v z + w z = u x + v x, u y + v y, u z + v z + w x, w y, w z = u x, u y, u z + v x, v y, v z + w x, w y, w z = u + v + w ck u = ck u x, u y, u z = ck u x, ck u y, ck u z = c ku x, c ku y, c ku z = c ku x, ku y, ku z = c ku k u + v = k u x, u y, u z + v x, v y, v z = k u x + v x, u y + v y, u z + v z

3 = k u x + v x, k u y + v y, k u z + v z = ku x + kv x, ku y + kv y, ku z + kv z = ku x, ku y, ku z + kv x, kv y, kv z = ku + kv e) u k + c = u x, u y, u z k + c = u x k + c, u y k + c, u z k + c = ku x + cu x, ku y + cu y, ku z + cu z = ku x, ku y, ku z + cu x, cu y, cu z = ku + cu 4. Solve the equation 1,, 3 x, 0, 4 = 1,, 3 for x. Use vector algebra to solve for x: 1,, 3 x, 0, 4 = 1,, 3, 4, 6 x +, 0, 4 =, 4, 6 x +, 0, 4 = 4, 8, 1 x = 6, 8, 8 x = 3, 4, 4 5. Let u = 1, 3, and v = 3, 4, 1. Normalize u and v. u = = = 14 u = u u = 1 14, 3 14, 14 v = = = 6 v = v v = 3 6, 4 6, 1 6. Let k be a scalar and let u = u x, u y, u z. Prove that ku = k u. 6 ku = ku x + ku y + ku z = k u x + u y + u z = k u x + u y + u z = k u 7. Is the angle between u and v orthogonal, acute, or obtuse?

4 a) u = 1, 1, 1, v =, 3, 4 b) u = 1, 1, 0, v =,, 0 c) u = 1, 1, 1, v = 3, 1, 0 a) u v = = 9 > 0 acute b) u v = = 0 orthogonal c) u v = = 4 < 0 obtuse 8. Let u = 1, 3, and v = 3, 4, 1. Find the angle θ between u and v. Using the equation u v = u v cos θ we have: u v 1 θ = cos u v = cos = cos 14 6 = Let u = u x, u y, u z, v = v x, v y, v z, and w = w x, w y, w z. Also let c and k be scalars. Prove the following dot product properties. a) u v = v u b) u v + w = u v + u w c) k u v = ku v = u kv d) v v = v e) 0 v = 0 u v = u x, u y, u z v x, v y, v z = u x v x + u y v y + u z v z = v x, v y, v z u x, u y, u z = v u u v + w = u x, u y, u z v x + w x, v y + w y, v z + w z = u x v x + w x + u y v y + w y + u z v z + w z = u x v x + u x w x + u y v y + u y w y + u z v z + u z w z = u x v x + u y v y + u z + u x w x + u y w y + u z w z = u v + u w k u v = k u x v x + u y v y + u z v z = ku x v x + ku y v y + ku z v z

5 = ku v = u x kv x + u y kv y + u z kv z = u kv v v = v x v x + v y v y + v z v z = v x + v y + v z = v 0 v = 0v x + 0v y + 0v z = Use the law of cosines (c = a + b ab cos θ, where a, b, and c are the lengths of the sides of a triangle and θ is the angle between sides a and b) to show u x v x + u y v y + u z v z = u v cos θ Hint: Consider Figure 1.9 and set c = u v, a = u and b = v, and use the dot product properties from the previous exercise. c = a + b ab cos θ u v = u + v u v cos θ u v u v = u u + v v u v cos θ u u u v + v v = u u + v v u v cos θ u v = u v cos θ 11. Let n =, 1. Decompose the vector g = 0, 9.8 into the sum of two orthogonal vectors, one parallel to n and the other orthogonal to n. Also, draw the vectors relative to a D coordinate system. g = proj n g = g n 9.8 n = n 5,1 = 1.96,1 = 3.9, 1.96 g = g g = 0, , 1.96 = 3.9, Let u =, 1, 4 and v = 3, 4, 1. Find w = u v, and show w u = 0 and w v = 0. w = u v = u y v z u z v y, u z v x u x v z, u x v y u y v x

6 = , 1 +, 8 3 = 17, 14, 5 w u = = = 0 w v = = = Let the following points define a triangle relative to some coordinate system: A = 0, 0, 0, B = 0, 1, 3, and C = 5, 1, 0. Find a vector orthogonal to this triangle. Hint: Find two vectors on two of the triangle s edges and use the cross product. u = B A = 0, 1, 3 v = C A = (5,1,0) n = u v = u y v z u z v y, u z v x u x v z, u x v y u y v x = 0 3, 15 0, 0 5 = ( 3, 15, 5) 14. Prove that u v = u v sin θ. Hint: Start with u v sin θ and use the trigonometric identity cos θ + sin θ = 1 sin θ = 1 cos θ; then apply Equation 1.4. To make the derivation simpler, we compute the following three formulas up front: u v = u x + u y + u z v x + v y + v z = u x v x + v y + v z + u y v x + v y + v z + u z v x + v y + v z = u x v x + u x v y + u x v z + u y v x + u y v y + u y v z + u z v x + u z v y + u z v z (1) u v = u x v x + u y v y + u z v z u x v x + u y v y + u z v z = u x v x u x v x + u y v y + u z v z + u y v y u x v x + u y v y + u z v z + u z v z u x v x + u y v y + u z v z = u x v x u x v x + u x v x u y v y + u x v x u z v z + u y v y u x v x + u y v y u y v y + u y v y u z v z + u z v z u x v x + u z v z u y v y + u z v z u z v z = u x v x + u x v x u y v y + u x v x u z v z + u y v y + u y v y u z v z + u z v z () u v u v = u x v y + u x v z + u y v x + u y v z + u z v x + u z v y u x v x u y v y u x v x u z v z u y v y u z v z (3)

7 = u y v z u y v y u z v z + u z v y + u z v x u x v x u z v z + u x v z + u x v y + u x v x u y v y + u y v x Now, u v sin θ = u v 1 cos θ = u v 1 = u v u v u v u v u v u v = u v u v And u v = u y v z u z v y, u z v x u x v z, u x v y u y v x = u y v z u z v y + u z v x u x v z + u x v y u y v x = u y v z u y v z u z v y + u z v y + u z v x u z v x u x v z + u x v z + u x v y u x v y u y v x + u y v x = u v u v Thus we obtain the desired result: u v = u v sin θ 15. Prove that u v gives the area of the parallelogram spanned by u and v; see Figure below. The area is the base times the height: A = v h

8 Using trigonometry, the height is given by h = u sin (θ). This, along with the application of Exercise 14, we can conclude: A = u v sin (θ) = u v 16. Give an example of 3D vectors u, v, and w such that u v w u v w. This shows the cross product is generally not associative. Hint: Consider combinations of the simple vectors i = 1, 0, 0, j = 0, 1, 0, and k = 0, 0, 1. Choose u = (1,1,0), v = i, and w = j. Then: But, u v w = 1,1,0 k = (1, 1,0) u v w = k j = i 17. Prove that the cross product of two nonzero parallel vectors results in the null vector; that is, u ku = 0. Hint: Just use the cross product definition. u ku = u y ku z u z ku y, u z ku x u x ku z, u x ku y u y ku x = ku y u z ku z u y, ku z u x ku x u z, ku x u y ku y u x = 0