Understanding Engineering Mathematics

x

O

y

z P(x,y,z)

OPr

Figure 11.12Position vector of a pointP.

Themagnitudeorlengthof a vectora=a 1 i+a 2 j+a 3 kis given by

a=|a|=

√

a^21 +a^22 +a 32

(also calledmodulusornormofa).
Aunit vectoris one of unit norm or magnitude – usually denoted with a circumflex,
for exampleaˆ.

Problem 11.4

Show that a=

√

3

2

iY

1

2

jisaunitvector.

We have|a|=

√√

√

√

(√

3

2

) 2

+

(

1

2

) 2

+ 02 =1.

Problem 11.5

Construct a unit vector parallel to a=iYj−3k.

To construct a unit vector parallel to a given vector we have only to divide by its modulus,
which fora=i+j− 3 kis

|a|=

√

12 + 12 +(− 3 )^2 =

√

11

so a unit vector parallel toais

ˆa=

1

√

11

(i+j− 3 k)

In general a vectora=a 1 i+a 2 j+a 3 kismultiplied by a scalarλas follows:

λa=λa 1 i+λa 2 j+λa 3 k

i.e. all components are multiplied byλ.
Note that the magnitude ofλais:

|λa|=λ|a|=λa