082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

7.6 The vector product 247
a 3 b
u
b
Plane containing
a and b
a
b
Figure7.29
a ×b is perpendicular to the plane containing
aand b.Theright-handed screw ruleallows
the direction ofa ×b to be found.
a
Figure7.30
Right-handedscrew ruleallowsthe
directionofb ×ato be found.
where ê is the unit vector required to define the appropriate direction, that is ê is a unit
vectorperpendicular toaandtobinasensedefined bytheright-handedscrewrule.To
evaluate b ×a we must imagine turning the screw from the direction of b towards that
of a. The screwwill advance as shown inFigure 7.30.
We notice immediately that a ×b ≠ b ×a since their directions are different. From
the definition of the vector product, itispossible toshow thatthe following rules hold:
a ×b = −(b ×a)
a × (b +c) = (a ×b) + (a ×c)
k(a ×b) = (ka) ×b =a× (kb)
the vector product is notcommutative
thedistributiverule
wherekisascalar
Example7.18 Ifaandbare parallel show thata ×b = 0.
Solution If a and b are parallel then the angle between a and b is zero. Therefore, a × b =
|a‖b|sin0ê= 0.Notethattheanswerisstillavector,andthatwedenotethezerovector
0i +0j +0k by0, todistinguish itfromthe scalar 0.Inparticular, wenote that
i×i=j×j=k×k=0
Ifaandbareparallel,thena ×b = 0.
Inparticular:
i×i=j×j=k×k=0
Example7.19 Show thati ×j = kand find expressions forj ×kandk ×i.
Solution We note that i and j are perpendicular so that |i ×j| = |i‖j|sin90 ◦ = 1. Furthermore,
theunitvectorperpendiculartoiandtojinthesensedefinedbytheright-handedscrew
rule is k. Therefore, i ×j = k as required. Similarly you should be able to show that
j×k=iandk×i=j.
i×j=k j×k=i k×i=j
j×i=−k k×j=−i i×k=−j