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

246 Chapter 7 Vectors
10 Find the modulusofa = i −j−k,the modulusof
b = 2i +j+2kand the scalar producta ·b.Deduce
the angle betweenaand b.
11 Ifa=2i+2j−kandb=3i−6j+2k,find
|a|, |b|,a·b and the angle betweenaand b.
12 Useavector methodto showthat the diagonals ofthe
rhombusshown in Figure7.28 intersect at90 ◦ .
A
B
D
Figure7.28
TherhombusABCD.
C
13 Usethe scalar product to findatwo-dimensional
vector a =a 1 i +a 2 j perpendicular to the vector
4i −2j.
14 Ifa=3i−2j,b=7i+5jandc=9i−j,showthat
a·(b−c)=(a·b)−(a·c).
15 Findthe work done bythe forceF = 3i −j+k
in moving an object through adisplacement
s=3i+5j.
16 Aforce ofmagnitude 14Nactsin the direction
i +j+k upon an object. Itcauses the object to move
from point A(2,1,0)to point B(3,3,3).Find the
work done bythe force.
17 (a) Usethe scalar product to showthat the
component ofain the direction ofbisa · ˆb,
where ˆb isaunitvector in the directionofb.
(b) Findthe component of2i +3j in the directionof
i+5j.
Solutions
1 −22, −22,58,20
2 (a) 21 (b) 3 (c) 7 (d) 6
3 19
4 47.62 ◦
5 82.11 ◦
6 18.4 ◦
7 7
8 14,16, 26
9 2|q| 2 ,7|q| 2 , −2|q| 2
10 √ 3,3, −1,101.1 ◦
11 3,7, −8,112.4 ◦
13 c(i +2j),cconstant
15 4J
16 48.5 J
17 17/ √ 26
7.6 THEVECTORPRODUCT
The result of finding the vector product of a and b is a vector of length |a||b|sinθ,
where θ is the angle between a and b. The direction of this vector is such that it is
perpendicular to a and to b, and so it is perpendicular to the plane containing a and
b (Figure 7.29). There are, however, two possible directions for this vector, but it is
conventionaltochoosetheoneassociatedwiththeapplicationoftheright-handedscrew
rule. Imagine turning a right-handed screw in the sense from a towards b as shown. A
right-handedscrewisonewhich,whenturnedclockwise,entersthematerialintowhich
it is being screwed. The direction in which the screw advances is the direction of the
required vector product. The symbol we shall use to denote the vector product is ×.
Formally, wewrite
a×b=|a‖b|sinθê