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

236 Chapter 7 Vectors
theoremOBhaslength √ x 2 +y 2 .Then,applyingPythagoras’stheoremtoright-angled
triangle OBP which has perpendicular sides OBand BP =z, we find
|r|=OP=
as required.
√
OB 2 +BP 2
= √ (x 2 +y 2 )+z 2
= √ x 2 +y 2 +z 2
Ifr=xi+yj+zk then |r|= √ x 2 +y 2 +z 2
Inthreedimensions we have the following general result:
Given vectors a = → OA=a 1
i+a 2
j+a 3
kandb= → OB =b 1
i+b 2
j+b 3
k,then
and
→
AB=b−a=(b 1
−a 1
)i+(b 2
−a 2
)j+(b 3
−a 3
)k
| AB|=|b−a|=|(b → 1
−a 1
)i+(b 2
−a 2
)j+(b 3
−a 3
)k|
√
= (b 1
−a 1
) 2 +(b 2
−a 2
) 2 +(b 3
−a 3
) 2
Example7.9 Ifa = 3i −2j +k andb = −2i +j−5k, find
(a) |a| (b) â (c) |b| (d) ˆb (e)b−a (f)|b−a|
Solution (a) |a| = √ 3 2 + (−2) 2 +1 2 = √ 14
(b) â = a
|a| = 1 √
14
(3i−2j+k)= 3 √
14
i − 2 √
14
j + 1 √
14
k
(c) |b| = √ (−2) 2 +1 2 + (−5) 2 = √ 30
(d) ˆb = b
|b| = √ 1 (−2i+j−5k)= √ −2 i + √ 1 j − √ 5 k
30 30 30 30
(e) b−a=−5i+3j−6k
(f) |b−a|= √ (−5) 2 +3 2 + (−6) 2 = √ 70
7.3.1 Thezerovector
A vector, all the components of which are zero, is called a zero vector and is denoted
by 0 to distinguish it from the scalar 0. Clearly the zero vector has a length of 0; it is
unusualinthatithas arbitrarydirection.