Game Development Reference
In-Depth Information
2
4
3
5
2
4
3
5
2
4
3
5
2
4
−1−0
3
5
2
4
−1
3
5
−1
0
0
0
1
(−1)(1)−(0)(0)
(0)(0)−(0)(1)
(0)(0)−(−1)(0)
13.
(a)
×
=
=
0−0
0−0
=
0
0
2
3
2
3
2
3
2
3
2
3
0
0
1
−1
0
(0)(0)−(1)(−1)
(1)(0)−(0)(0)
(0)(−1)−(0)(0)
0−(−1)
0−0
0−0
1
0
0
4
5
4
5
4
5
4
5
4
5
×
=
=
=
2
4
−2
3
5
2
4
3
5
2
4
3
5
2
4
−4−(−2)
3
5
2
4
−
−1
0
3
5
−
−1
(4)(−1)−(1)(−2)
(1)(1)−(−2)(−1)
(−2)(−2)−(4)(1)
(b)
4
1
×
=
=
1−2
4−4
=
2
3
2
4
−2
3
2
3
2
4
−2−(−4)
3
2
3
−
−1
(−2)(1)−(−1)(4)
(−1)(−2)−(1)(1)
(1)(4)−(−2)(−2)
2
1
0
4
5
5
4
5
5
4
5
×
4
1
=
=
2−1
4−4
=
2
4
3
5
2
4
3
5
2
4
3
5
2
4
3
5
2
4
3
5
3
10
7
−7
4
(10)(4)−(7)(−7)
(7)(8)−(3)(4)
(3)(−7)−(10)(8)
40−(−49)
56−12
−21−80
89
44
−101
(c)
×
=
=
=
2
4
3
5
2
4
3
5
2
4
3
5
2
4
−49−40
3
5
2
4
−89
3
5
−7
4
3
10
7
(−7)(7)−(4)(10)
(4)(3)−(8)(7)
(8)(10)−(−7)(3)
×
=
=
12−56
80−(−21)
=
−44
101
2
4
3
5
2
4
3
5
a
x
a
y
a
z
b
x
b
y
b
z
14. Let a =
and b =
. Then ab = abcos θ and a×b =
2
4
3
5
a
y
b
z
−a
z
b
y
a
z
b
x
−a
x
b
z
a
x
b
y
−a
y
b
x
. From a×b, we have:
a×b =
(a
y
b
z
−a
z
b
y
)
2
+ (a
z
b
x
−a
x
b
z
)
2
+ (a
x
b
y
−a
y
b
x
)
2
=
a
y
b
z
−2a
y
a
z
b
y
b
z
+ a
z
b
y
+ a
z
b
x
−2a
x
a
z
b
x
b
z
+ a
x
b
z
+ a
x
b
y
−2a
x
a
y
b
x
b
y
+ a
y
b
x
.
If we now consider absin θ, we find that:
absin θ = ab
1−cos
2
θ
2
a
x
b
x
+ a
y
b
y
+ a
z
b
z
=
a
x
+ a
y
+ a
z
b
x
+ b
y
+ b
z
1−
a
x
+ a
y
+ a
z
b
x
+ b
y
+ b
z
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