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or
+
=
−
i
m
q
d
(9.6)
We again define
q
in terms of
p
,sowelet
q
=
p
where is a scalar that requires defining.
As
q
and
p
have the same projection on
n
, we have
n
·
q
=
n
·
p
and
·
·
n
d
d
n
k
=
p
=
p
We now define
n
in terms of
ijk
. Therefore, from Eq. (9.5), we have
n
n
·
n
·
=−
sin
j
+
cos
k
and
d cos
z
P
cos
Returning to Eq. (9.6), we substitute
m
in terms of
ijk
.
From Eq. (9.5), we have
=
−
y
P
sin
+
m
=
cos
j
+
sin
k
and Eq. (9.6) becomes
i
+
cos
j
+
sin
k
=
x
P
i
+
y
P
j
+
z
P
k
−
d
k
(9.7)
We now isolate the
i
components of Eq. (9.7) to reveal :
=
x
P
Finally, substituting gives
dx
P
cos
z
P
cos
We isolate the
j
components of Eq. (9.7) to reveal :
=
−
y
P
sin
+
y
P
cos
=
and substituting gives
y
P
cos
d cos
=
−
y
P
sin
+
z
P
cos
dy
P
=
z
P
cos
−
y
P
sin
+
Note that when
0, and equal the values for the orthogonal projection plane.
Having examined two orientations of the projection plane, let's complete this analysis with
an arbitrary orientation of the plane.
=
