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

234 Chapter 7 Vectors
y
5
B(–3, 2)
4
3
2
b
1
–6 –5 –4 –3 –2 –1 0
a
1 2 3 4
A(5, 4)
5 6 x
Figure7.20
Points Aand Bin thex--yplane.
Therefore,
→
AB=b−a
= (−3i +2j) − (5i +4j)
=−8i−2j
Alternatively, intermsof column vectors
b−a=
=
( ) ( −3 5
−
2 4)
( ) −8
−2
Wenotethatsubtraction(andlikewiseaddition)ofcolumnvectorsiscarriedoutcomponentbycomponent.Tofind
| → AB|wemustobtainthelengthofthevector → AB.Referring
toFigure7.20,wenotethatthisquantityisthelengthofthehypotenuseofaright-angled
triangle with perpendicular sides 8 and 2.Thatis, | → AB| = √ 8 2 +2 2 = √ 68 = 8.25.
More generallywehave the following result:
Given vectors a = → OA=a 1
i+a 2
jandb= → OB =b 1
i +b 2
j (Figure7.21), then
and
→
AB=b−a=(b 1
−a 1
)i+(b 2
−a 2
)j
| AB|=|b−a|=|(b → 1
−a 1
)i+(b 2
−a 2
)j|
√
= (b 1
−a 1
) 2 +(b 2
−a 2
) 2
Example7.7 Ifa =
( ( ) 7 −2
andb= ,
3)
5
(a) find a +b, a −b, b +aandb −a, commentingupon the results
(b) find 2a −3b
(c) find |a −b|.