Game Development Reference
In-Depth Information
1−
(a
x
b
x
+ a
y
b
y
+ a
z
b
z
)
2
=
a
x
+ a
y
+ a
z
b
x
+ b
y
+ b
z
a
x
+ a
y
+ a
z
b
x
+ b
y
+ b
z
=
a
x
+ a
y
+ a
z
b
x
+ b
y
+ b
z
−(a
x
b
x
+ a
y
b
y
+ a
z
b
z
)
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
.
Starting from both ends, we have met in the middle, proving that
a×b = absin θ.
15.
(a) (1)
3
4
1
= |3|+|4| = 7
|3|
2
+|4|
2
= 5
3
4
2
=
√
91 ≈ 4.498
|3|
3
+|4|
3
=
3
3
3
4
3
=
3
4
∞
= max (|3|,|4|) = 4
(2)
5 −12
1
= |5|+|−12| = 17
|5|
2
+|−12|
2
= 13
5 −12
2
=
√
|5|
3
+|−12|
3
=
3
3
5 −12
3
=
1853 ≈ 12.283
5 −12
∞
= max (|5|,|−12|) = 12
(3)
−2
10 −7
1
= |−2|+|10|+|−7| = 19
√
153 ≈ 12.369
|−2|
2
+|10|
2
+|−7|
2
=
−2
10 −7
2
=
√
|−2|
3
+|10|
3
+|−7|
3
=
−2
10 −7
3
=
3
3
1351 ≈ 11.055
−2
10 −7
∞
= max (|−2|,|10|,|−7|) = 10
(4)
6
1 −9
1
= |6|+|1|+|−9| = 16
√
|6|
2
+|1|
2
+|−9|
2
=
6
1 −9
2
=
118 ≈ 10.863
√
946 ≈ 9.817
|6|
3
+|1|
3
+|−9|
3
=
6
1 −9
3
3
3
=
6
1 −9
∞
= max (|6|,|1|,|−9|) = 9
(5)
−2 −2 −2 −2
1
= |−2|+|−2|+|−2|+|−2| = 8
|−2|
2
+|−2|
2
+|−2|
2
+|−2|
2
= 4
−2 −2 −2 −2
2
=
√
|−2|
3
+|−2|
3
+|−2|
3
+|−2|
3
=
3
3
−2 −2 −2 −2
3
=
32 ≈ 3.175
−2 −2 −2 −2
∞
= max (|−2|,|−2|,|−2|,|−2|) = 2
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