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

7.3 Cartesian components 235
Solution (a)
a 2
O
b
a
a 1
A
b – a
b 1
Figure7.21
Thequantity √
|b−a|= (b 1 −a 1 ) 2 + (b 2 −a 2 ) 2 .
a+b=
a−b=
b+a=
b−a=
( ( ) ( 7 −2 5
+ =
3)
5 8)
x
( ( 7 −2
− =
3)
5)
( ( −2 7
+ =
5)
3)
( ) 9
−2
( 5
8)
( ( ( −2 7 −9
− =
5)
3)
2)
A
x
z
O
{
P}
z
r
B
}
y
Figure7.22
Thequantity |r| = √ x 2 +y 2 +z 2 .
x
y
y
b 2
B
We note thataddition is commutative whereas subtraction isnot.
( ( ( ( ) ( )
7 −2 14 −6 20
(b) 2a−3b=2 −3 = − =
3)
5)
6)
15 −9
(c) |a−b|=|9i−2j|= √ 9 2 + (−2) 2 = √ 85
The previous development readily generalizes to the three-dimensional case. Taking
Cartesian axes x, y and z, any point in three-dimensional space can be represented by
givingx,yandzcoordinates(Figure7.22).Denotingunitvectorsalongtheseaxesbyi,
jandk, respectively, wecan write the vectorfrom OtoP(x,y,z) as
→
OP=r=xi+yj+zk=
⎛ ⎞
x
⎝y⎠
z
Thevectors i,jand kareorthogonal.
Example7.8 Ifr =xi +yj +zk show thatthe modulus ofrisr = √ x 2 +y 2 +z 2 .
Solution Recalling Figure 7.22 we first calculate the length of OB. Now OAB is a right-angled
triangle with perpendicular sides OA = x and AB = y. Therefore by Pythagoras’s