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and
c = i + j
Let's substitute the lines in the original equations:
L 1 : s =
i + j
. v ) 2
2 + r 2 =0
( s
s
2+1=
1
the negative discriminant confirms a miss condition.
L 2 : s =
i + j
. v ) 2
2 + r 2 =1
( s
s
2+1=0
the zero discriminant confirms a touch condition, therefore λ =1thetouch
point is P 2 (2 , 1 , 0) which is correct.
L 3 : s =
i + j
. v ) 2
2 + r 2 =2
( s
s
2+1=1
the positive discriminant confirms an intersect condition, therefore
1=1+ 2 r 2
2
2 ±
λ =
1
The interse ct ion points are given by the two values of λ :
if λ =1+ 2
x P =2+ 1+ 2
=1
1
2
1
2
y P =0+ 1+ 2
1
2 =1+
1
2
z P =0
if λ = 2
1
1
=1+
x P =1+ 2
1
2
1
2
y P =0+ 2
1
1
2 =1
1
2
z P =0
The intersection points are P 3 1
2 , 0 and P 3 1+
2 , 0
1
1
1
1
2 , 1+
2 , 1
which are correct.
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