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

242 Chapter 7 Vectors
From the definition of the scalar product, it is possible to show that the following
rules hold:
a·b = b·a the scalar product iscommutative
k(a·b) = (ka·b) wherekisascalar
(a +b)·c = (a·c) + (b·c) thedistributiverule
Itisimportantatthisstagetorealizethatnotationisveryimportantinvectorwork.You
should not use a× to denote the scalar product because this is the symbol we shall use
for the vector product.
Example7.12 If a and b are parallel vectors, show that a·b = |a‖b|. If a and b are orthogonal show
that theirscalar product iszero.
Solution If a and b are parallel then the angle between them is zero. Therefore a·b =
|a‖b|cos0 ◦ = |a||b|. If a and b are orthogonal, then the angle between them is 90 ◦
and a·b = |a||b|cos90 ◦ = 0.
Similarly we can show that if a and b are two non-zero vectors for which a·b = 0,
thenaandbmustbeorthogonal.
If a andbareparallel vectors, a·b = |a‖b|.
If a andbareorthogonal vectors,a·b = 0.
Animmediateconsequenceofthepreviousresultisthefollowingusefulsetofformulae:
i·i=j·j=k·k=1
i·j=j·k=k·i=0
Example7.13 Ifa =a 1
i+a 2
j+a 3
kandb =b 1
i+b 2
j+b 3
kshowthata·b =a 1
b 1
+a 2
b 2
+a 3
b 3
.
Solution We have
a·b = (a 1
i+a 2
j+a 3
k)·(b 1
i+b 2
j+b 3
k)
=a 1
i·(b 1
i+b 2
j+b 3
k)+a 2
j·(b 1
i+b 2
j+b 3
k)
+a 3
k·(b 1
i+b 2
j+b 3
k)
=a 1
b 1
i·i+a 1
b 2
i·j+a 1
b 3
i·k+a 2
b 1
j·i+a 2
b 2
j·j+a 2
b 3
j·k
+a 3
b 1
k·i+a 3
b 2
k·j+a 3
b 3
k·k
=a 1
b 1
+a 2
b 2
+a 3
b 3
as required. Thus, given two vectors in component form their scalar product is the sum
of the products of corresponding components.
The resultdeveloped inExample 7.13 isimportant and should bememorized:
Ifa =a 1
i+a 2
j+a 3
kandb =b 1
i+b 2
j+b 3
k,
thena·b =a 1
b 1
+a 2
b 2
+a 3
b 3
.