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Unformatted text preview: .9
9A
1
0
0
0
1 which gives a single matrix:
0
0.9
0
0
B 0 0.9
0
B
@0
0 0.9
0
0
0 0.1n 1 1
1C
C
1A
1 Marks:
d 4 The scene is to be viewed from a moving viewpoint speciﬁed by its position C
and a left handed viewing coordinate system { u, v, w }.
T B 0 0.9
0
B
@0
0 0.9
0
0
0 Transformations" 0.1n 1C
C
1A
1 Marks: d The scene is to be viewed from a moving viewpoint speciﬁed by its position C
and a left handed viewing coordinate system { u, v, w }. At one point in the animation the view direction is w = (1, 0, 0) T , and the
viewpoint is given by C = (20, 5, 10). Given that the view is in the horizontal
plane (v = (0, 1, 0) T ) ﬁnd the value of u.
The reference coordinate system is assumed to be lefthanded. The viewing
coordinate system, { u, v, w }, is also lefthanded so we have the crossproduct
identities:
u⇥v = w
w⇥u = v
v⇥w = u 4 and a left handed viewing coordinate system { u, v, w }. At one point in the animation the view direction is w = (1, 0, 0) T , and the
viewpoint is given by C = (20, 5, 10). Given that the view is in the horizontal
plane (v = (0, 1, 0) T ) ﬁnd the value of u. Transformations" The reference coordinate system is assumed to be lefthanded. The viewing
coordinate system, { u, v, w }, is also lefthanded so we have the crossproduct
Department of Computing Examinations – 2011  2012 Session
Conﬁdential
identities:
u⇥v = w
MODEL ANSWERS and MARKING SCHEME
w⇥u = v
First Examiner: Paul Aljabar
Second Examiner: Daniel Rueckert
v⇥w = u
Paper: C317  Graphics
Question: 1
Page 4 of 16 The third identity can be used to ﬁnd the answer:
01 01 0 1
0
1
0
u = @1A ⇥ @0A = @ 0A
0
0
1 Marks:
e Hence, or otherwise, ﬁnd the viewing transformation matrix.
The viewing matrix has the form 4 u= 1
0 ⇥ 0
0 = 0
1 Marks:
Transformations"
e Hence, or otherwise, ﬁnd the viewing transformation matrix.
The viewing matrix has the form
0
u x uy uz
B v x vy vz
B
@ w x wy wz
000 1 C·u
C · vC
C
C · wA
1 u, v and w are known from the previous question.
C·u = 0 10 1 20
0
@ 5A · @ 0A = 10
10
1
0 1 01
20
0
@ A @A e Hence, or otherwise, ﬁnd the viewing transformation matrix.
The viewing matrix has the form
0
u x uy uz
B v x vy vz
B
@ w x wy wz
0
00 1
C·u
C · vC
C
C · wA
1 u, v and w are known from the previous question.
0 10 1
20
0
C · u = @ 5A · @ 0A = 10
10
1
0 1 01
20
0
C · v = @ 5A · @1A = 5
10
0
0 10 1
20
1
C · w = @ 5A · @ 0A = 20
10
0 Putting these together gives the viewing matrix as
0
1
00
1 10
B01
0 5C
B
C
@10
0 20A
00
01 Marks: 4...
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This document was uploaded on 03/26/2014.
 Spring '14

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