That is,
is a planar dyadic, which may be expressed by the equation
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(See No. 140.) Let
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the equation becomes
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or,
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or,
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This may be written
[1]
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Now if the dyadic
is given in any form, the scalars
![{\displaystyle \Phi _{\text{S}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51e8b4daf024dd54d1ed674fd6fb2e2deafdb6f2) ![{\displaystyle \{\Phi ^{-1}\}_{\text{S}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd2b1eaae7569182b895d5d3ae22e2dc6a5f388f)
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are easily determined We have therefore a cubic equation in
for which we can find at least one and perhaps three roots. That is, we can find at least one value of
and perhaps three, which will satisfy the equation
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By substitution of such a value,
becomes a planar dyadic, the planes of which may be easily determined.[2] Let
be a vector normal to the plane of the consequents. Then
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If
is a tonic, we may obtain three equations of this kind, say
![{\displaystyle \Phi .\alpha =a\alpha ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18023d0220798b7704a8b31c047e4bf0396384d3) ![{\displaystyle \Phi .\beta =b\beta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9c30c4496133dc240743b385e91be26257ca609)
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in which
are not complanar. Hence (by No. 108),
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where
are the reciprocals of
In any case, we may suppose
to have the same sign as
since the cubic equation must have such a root. Let
(as before) be normal to the plane of the consequents of the planar
and
normal to the plane of the antecedents, the lengths of
and
being such that
[3] Let
be any vector normal to
and such that
is not parallel to
(The case in which
is always parallel to
if
is perpendicular to
is evidently that of a tonic, and needs no farther discussion.)
and therefore
will be perpendicular to
The same will be true of
Now (by No. 140)
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that is,
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- ↑ [See note on p. 90.]
- ↑ In particular cases,
may reduce to a linear dyadic, or to zero. These, however, will present no difficulties to the student.
- ↑ For the case in which the two planes are perpendicular to each other, see No. 157.