Game Development Reference
In-Depth Information
4.
(a) First, let's find the normal by using Equation (9.12):
2
4
3
5
2
4
3
5
2
4
−3
3
5
−1
17
6
1
−2
e
3
= p
2
−p
1
=
−
=
−11
19
,
2
4
−9
3
5
2
4
3
5
2
4
−12
3
5
−1
17
e
1
= p
3
−p
2
=
8
0
−
=
−17
,
2
4
3
5
2
4
3
5
2
4
3
5
(−11)(−17)−(19)(9)
(19)(−12)−(−3)(−17)
(−3)(9)−(−11)(−12)
187−171
−228−51
−27−132
16
−279
−159
e
3
×e
1
=
=
=
.
Let's normalize it:
√
e
3
×e
1
=
16
2
+ (−279)
2
+ (−159)
2
=
103378 ≈ 321.5,
16 −279 −159
e
3
×e
1
e
3
×e
1
≈
n =
321.5
≈
.04976 −.8677 −.4945
.
Just for kicks, we'll verify that we get the same result with Equation (9.13) from
Section 9.5.3:
n
x
= (z
1
+ z
2
)(y
1
−y
2
) + (z
2
+ z
3
)(y
2
−y
3
) + (z
3
+ z
1
)(y
3
−y
1
)
= ((−2) + 17)(10−(−1)) + (17 + 0)((−1)−8) + (0 + (−2))(8−10)
= 16,
n
y
= (x
1
+ x
2
)(z
1
−z
2
) + (x
2
+ x
3
)(z
2
−z
3
) + (x
3
+ x
1
)(z
3
−z
1
)
= (6 + 3)((−2)−17) + (3 + (−9))(17−0) + ((−9) + 6)(0−(−2))
= −279,
n
z
= (y
1
+ y
2
)(x
1
−x
2
) + (y
2
+ y
3
)(x
2
−x
3
) + (y
3
+ y
1
)(x
3
−x
1
)
= (10 + (−1))(6−3) + ((−1) + 8)(3−(−9)) + (8 + 10)((−9)−6)
= −159,
.04976 −.8677 −0.4945
16 −279
159
16 −279
159
n =
≈
≈
.
321.5
16
2
+ (−279)
2
+ 159
2
Now that we have n, we can compute d. We'll arbitrarily use p
1
:
d = np
1
≈
.04976 −.8677 −.4945
6
10 −2
≈ (.04976)(6) + (−.8677)(10) + (−.4945)(−2) ≈−7.389.
The plane equation for this triangle is
.04976x−.8677y −.4945z = −7.389.
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