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

7.5 The scalar product 243
Example7.14 Ifa = 5i −3j +2kand b = −2i +4j +kfind the scalar producta·b.
Solution Usingthe previous resultwe find
a·b=(5i−3j+2k)·(−2i+4j+k)
= (5)(−2) + (−3)(4) + (2)(1)
=−10−12+2
= −20
Example7.15 Ifa =a 1
i +a 2
j +a 3
k, find
(a) a·a (b) |a| 2
Solution (a) Using the previous resultwefind
a·a = (a 1
i+a 2
j+a 3
k)·(a 1
i+a 2
j+a 3
k)
=a 2 1 +a2 2 +a2 3
(b) From Example 7.8 we know that the modulus ofr =xi +yj +zk is √ x 2 +y 2 +z 2
and therefore
√
|a| = a 2 1 +a2 2 +a2 3
so that
|a| 2 =a 2 1 +a2 2 +a2 3
We notethe generalresultthat
a·a=|a| 2
Example7.16 Ifa = 3i +j−k andb = 2i +j+2k finda·band the angle betweenaandb.
Solution We have
a·b = (3)(2) + (1)(1) + (−1)(2)
=6+1−2
= 5
Furthermore, from the definition ofthe scalar producta·b = |a||b|cos θ. Now,
|a| = √ 9+1+1= √ 11 and |b| = √ 4+1+4=3
Therefore,
cosθ = a·b
|a‖b| = 5
3 √ 11
fromwhich wededuce that θ = 59.8 ◦ or 1.04 radians.