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where is a scalar that requires defining.
As
q
and
p
have the same projection on
n
, then
n
·
q
=
n
·
p
and
n
·
d
d
n
·
k
p
We now define
n
in terms of
ijk
. Therefore, from Eq. (9.2), we find that
=
p
=
n
·
n
·
n
=
sin
i
+
cos
k
and
d cos
x
P
sin
z
P
cos
Returning to Eq. (9.3), we substitute
l
and
m
in terms of
ijk
.
From Eq. (9.2), we have
=
+
l
=
cos
i
−
sin
k
m
=
j
and Eq. (9.3) becomes
cos
i
−
sin
k
+
j
=
x
P
i
+
y
P
j
+
z
P
k
−
d
k
(9.4)
We now isolate the
i
components of Eq. (9.4) to reveal :
cos
=
x
P
and
x
P
cos
=
Finally, substituting gives
x
P
cos
d cos
x
P
sin
=
+
z
P
cos
dx
P
x
P
sin
z
P
cos
We now isolate the
j
components of Eq. (9.4) to reveal :
=
+
=
y
P
and substituting gives
dy
P
cos
x
P
sin
=
z
P
cos
+
Thus, we have found Q . Note that when
=
0, and equal the values for the orthogonal
projection plane.
