Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/105

component is the same in both, but the major axes of the ellipses are perpendicular. The case in which the directions of ${\mathfrak {r}}$ and ${\mathfrak {s}}$ are real, forms no exception to this rule.

It will be observed that every circular bivector is perpendicular to itself, and to every parallel bivector.

If two bivectors, $\mu +\iota \nu ,\,\mu '+\iota \nu ',$ which do not lie in the same plane are perpendicular, we may resolve $\mu$ and $\nu$ into components parallel and perpendicular to the plane of $\mu '$ and $\nu '.$ The components perpendicular to the plane evidently contribute nothing to the value of

$[\mu +\iota \nu ].[\mu '+\iota \nu ']$

Therefore the components of $\mu$ and $\nu$ parallel to the plane of $\mu ',\nu ',$ form a bivector which is perpendicular to $\mu '+\iota \nu '.$ That is, if two bivectors are perpendicular, the directional ellipse of either, projected upon the plane of the other and rotated through a quadrant in that plane, will be similar and similarly situated to the directional ellipse of the second.

8. A bivector may be divided in one and only one way into parts parallel and perpendicular to another, provided that the second is not circular. If ${\mathfrak {a}}$ and ${\mathfrak {b}}$ are the bivectors, the parts of ${\mathfrak {a}}$ will be

${\frac {\mathfrak {a.b}}{\mathfrak {bb}}}{\mathfrak {b}}$⁠and⁠${\mathfrak {a}}-{\frac {\mathfrak {b.a}}{\mathfrak {b.b}}}{\mathfrak {b}}.$

If ${\mathfrak {b}}$ is circular, the resolution of ${\mathfrak {a}}$ is impossible, unless it is perpendicular to ${\mathfrak {b}}.$ In this case the resolution is indeterminate.

9. Since ${\mathfrak {a}}\times {\mathfrak {b}}.{\mathfrak {a}}=0,$ and ${\mathfrak {a}}\times {\mathfrak {b}}.{\mathfrak {b}}=0,\,{\mathfrak {a}}\times {\mathfrak {b}}$ is perpendicular to ${\mathfrak {a}}$ and ${\mathfrak {b}}.$ We may regard the plane of the product as determined by the condition that the directional ellipses of the factors projected upon it become similar and similarly situated. The directional ellipse of the product is similar to these projections, but its orientation is different by 90°. It may easily be shown that ${\mathfrak {a}}\times {\mathfrak {b}}$ vanishes only with ${\mathfrak {a}}$ or ${\mathfrak {b}},$ or when ${\mathfrak {a}}$ and ${\mathfrak {b}}$ are parallel.

10. The bivector equation

$({\mathfrak {a}}\times {\mathfrak {b}}.{\mathfrak {c}}){\mathfrak {d}}-({\mathfrak {b}}.{\mathfrak {c}}\times {\mathfrak {d}}){\mathfrak {a}}+({\mathfrak {c}}.{\mathfrak {d}}\times {\mathfrak {a}}){\mathfrak {b}}-({\mathfrak {d}}.{\mathfrak {a}}\times {\mathfrak {b}}){\mathfrak {c}}=0$

is identical, as may be verified by substituting expressions of the form $xi+yj+zk$ ($x,y,z$ being biscalars), for each of the bivectors. (Compare No. 37.) This equation shows that if the product ${\mathfrak {a}}\times {\mathfrak {b}}$ of any two bivectors vanishes, one of these will be equal to the other with a biscalar coefficient, that is, they will be parallel, according to the definition given above. If the product ${\mathfrak {a}}.{\mathfrak {b}}\times {\mathfrak {c}}$ of any three bivectors vanishes, the equation shows that one of these may be expressed as a sum of the other two with biscalar coefficients. In this case, we may say (from the analogy of the scalar analysis) that the three bivectors are complanar. (This does not imply that they