Answers to Assignment 1 Qn 1
a)
����⃑ ����⃑
b)
The equation of the face afhd is −60𝑋𝑋 + 75𝑍𝑍 = 0 𝑖𝑖 𝑗𝑗 𝑘𝑘
fh
gb ac
de fgbhf
aceda
𝑖𝑖𝑗𝑗𝑘𝑘 (𝐴𝐴,𝐵𝐵,𝐶𝐶)=𝑎𝑎𝑎𝑎×𝑎𝑎𝑎𝑎=� 0 5 0 �=(−60,0,75)
−15 0 −12
(𝐴𝐴,𝐵𝐵,𝐶𝐶)=�𝑐𝑐��𝑐𝑐⃑×�𝑐𝑐��𝑐𝑐�⃑=�1 0 −12�=(60,0,5) 050
To solve for 60𝑋𝑋 + 5𝑍𝑍 + 𝐷𝐷 = 0 , we put (10,0,0) into the equation, 𝐷𝐷 = −600 The equation of the face cgbe is
60𝑋𝑋 + 5𝑍𝑍 − 600 = 0
The system of linear inequalities are
−60𝑋𝑋 + 75𝑍𝑍 < 0 60𝑋𝑋 + 5𝑍𝑍 − 600 < 0
1
−12 < 𝑍𝑍 < 0 0<𝑌𝑌<5
b) Super-ellipsoid
c) makes the sampling more uniform and avoid square/cubic root, which would cause part of the shape to become missing (either reason acceptable)
Qn 3
a)i) 𝑋𝑋2/𝑠𝑠 𝑌𝑌2/𝑠𝑠 𝑍𝑍2−[�2� 1+�2� 1]=1
𝑍𝑍 = sec (𝛼𝛼)
�𝑋𝑋2�2/𝑠𝑠1 + �𝑌𝑌2�2/𝑠𝑠1 = 𝑡𝑡𝑎𝑎𝑠𝑠2𝛼𝛼
Qn 2
a) 𝑍𝑍 = 4𝑠𝑠𝑖𝑖𝑠𝑠10 𝛼𝛼
𝑋𝑋 = 2𝑐𝑐𝑐𝑐𝑠𝑠10 𝛼𝛼 𝑐𝑐𝑐𝑐𝑠𝑠5𝛽𝛽 𝑌𝑌 = 2cos10 𝛼𝛼 𝑠𝑠𝑖𝑖𝑠𝑠5𝛽𝛽
1 �𝑋𝑋�1/𝑠𝑠1 = 𝑐𝑐𝑐𝑐𝑠𝑠𝛽𝛽 ⇒ 𝑋𝑋 = 2(𝑡𝑡𝑎𝑎𝑠𝑠𝑠𝑠 𝛼𝛼)(𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠 𝛽𝛽) tan(𝛼𝛼) 2 1 1
Similarly, 𝑌𝑌 = 2(𝑡𝑡𝑎𝑎𝑠𝑠𝑠𝑠1 𝛼𝛼)(𝑠𝑠𝑖𝑖𝑠𝑠𝑠𝑠1 𝛽𝛽)
a) ii)
Super-hyperboloid (Two-sheeted Super-hyperboloid) b)
cg
bf
2
cbfgc
daehd eh
ad
c)
����⃑
(𝐴𝐴,𝐵𝐵,𝐶𝐶) = 𝑎𝑎����𝑐𝑐⃑ × 𝑎𝑎𝑎𝑎 = (9,−9,−45) × (0,2,0) = �9 −9
5𝑋𝑋 + 𝑍𝑍 + 𝐷𝐷 = 0
Put in a(1, -1, -5) gives 𝐷𝐷 = 0. The plane passes through the origin. The plane equation is 5𝑋𝑋 + 𝑍𝑍 = 0
d)
By symmetry, PRP = (0, 0, 0)
𝑡𝑡𝑎𝑎𝑠𝑠 �𝜃𝜃2� = 15 ⟹ 𝜃𝜃 = 22.61986495
3
𝑖𝑖𝑗𝑗𝑘𝑘
−45� = (90,0,18) 020
The command is
gluPerspective (22.61986495, 1, 5, 50)