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from which
=
d
−ˆ
n
·
c
(6.36)
But from Eq. (6.34), we see that
−
CP
=
, we find
−
CP
and by finding
and the problem is solved.
Let's illustrate this technique with the following example: a plane is defined by
√
3
3
x
√
3
3
y
√
3
3
z
+
+
=
10
and as its normal vector is a unit vector, the perpendicular distance from the origin to the plane
is 10, as shown in Fig. 6.21.
Y
C
(
9
,9 ,9
)
Z
10
X
Figure 6.21.
The sphere's radius is 5 and its centre is located at 9 9 9, which makes it
√
243
15588
from the origin. Obviously, in this simple example we know that the sphere and plane do not
intersect.
Using Eq. (6.36), we find that
=
=
d
−ˆ
n
·
c
where
√
3
3
√
3
3
√
3
3
d
=
10
n
ˆ
=
i
+
j
+
k
c
=
9
i
+
9
j
+
9
k
then
9
√
3
3
9
√
3
3
9
√
3
3
= −
=
10
−
−
−
5588
=
5588
