Introdution to 3D Game Programming With DirectX11 第1章 習題解答

1.

(a) u + v = (1, 2) + (3, -4) = (4, -2)

(b) u – v = (1, 2) - (3, -4) = (-2, 6)

(c) 2u + 1/2v = 2(1, 2) + 1/2(3, -4) = (2,4) + (3/2, -2) = (7/2, 2)

(d) -2u + v = -2(1, 2) + (3, -4) = (-2, -4)+ (3, -4) = (1, -8)

2.

(a) u + v = (-1, 3, 2) + (3, -4, 1) = (2,-1, 3)

(b) u – v = (-1, 3, 2) – (3, -4, 1) = (-4,7, 1)

(c) 3u + 2v = 3(-1, 3, 2) + 2(3, -4, 1) =(-3, 9, 6) + (6, -8, 2) = (3, 1, 8)

(d) -2u +v = -2(-1, 3, 2) + (3, -4, 1) =(2, -6, -4) + (3, -4, 1) = (5, -10, -3)

3.

(a) u + v = (ux, uy,uz) + (vx, vy, vz) = (ux+vx,uy+vy, uz+vz) = (vx+ux,vy+uy, vz+uz) = (vx, vy,vz) + (ux, uy, uz) = v + u

(b) u + (v + w) = (ux, uy,uz) +((vx, vy, vz) + (wx,wy, wz)) = (ux+vx+wx, uy+vy+wy,uz+vz+wz) = ((ux, uy, uz)+(vx, vy, vz)) + (wx, wy,wz) = (u + v) + w

(c),(d),(e)略

4.略

5.

||u|| = (-1, 3, 2)/  = (-1, 3, 2)/  = (-1, 3, 2)/

||v|| = (3, -4, 1)/  = (3, -4, 1)/

6.略

7.略

8.略

9.略

10.

 =  +  – 2abcos

=  -2abcos

uxvx + uyvy+ uzvz = abcos  = ||a||||b||cos

11.

||n|| = (-2, 1)/  = (-2, 1)/  

12.

(-2, 1, 4)cross(3, -4, 1) = (1*1-(-4)*4, -(1*(-2) – 3*4), (-2)*(-4)-1*3)=(17,14, 5)

略

13.

AB cross AC = (0, -1, -3) cross (-5, -1, 0) = (-3, 15, -5)

14.

略

15.

同14

16.

略

17

略

18.

(1,0,0) (1,5,0) (-16,-8,32)

19.

略