Higher Engineering Mathematics, Sixth Edition

278 Higher Engineering Mathematics

(i) From equation (2),

if p=a 1 i+a 2 j+a 3 k

and q=b 1 i+b 2 j+b 3 k

then p•q=a 1 b 1 +a 2 b 2 +a 3 b 3

When p= 2 i+j−k,

a 1 = 2 ,a 2 =1anda 3 =− 1

and when q=i− 3 j+ 2 k,

b 1 = 1 ,b 2 =−3andb 3 = 2

Hence p•q=( 2 )( 1 )+( 1 )(− 3 )+(− 1 )( 2 )

i.e. p•q=− 3

(ii) p+q=( 2 i+j−k)+(i− 3 j+ 2 k)

= 3 i− 2 j+k

(iii) |p+q|=| 3 i− 2 j+k|

From equation (3),

|p+q|=

√

[3^2 +(− 2 )^2 + 12 ]=

√

14

(iv) From equation (3),

|p|=| 2 i+j−k|

=

√

[2^2 + 12 +(− 1 )^2 ]=

√

6

Similarly,

|q|=|i− 3 j+ 2 k|

=

√

[1^2 +(− 3 )^2 + 22 ]=

√

14

Hence|p|+|q|=

√

6 +

√

14 =6.191, correct to 3

decimal places.

Problem 4. Determine the angle between vectors

oaandobwhen

oa=i+ 2 j− 3 k

and ob= 2 i−j+ 4 k.

An equation for cosθis given in equation (4)

cosθ=

a 1 b 1 +a 2 b 2 +a 3 b 3

√

(a 12 +a 22 +a 32 )

√

(b 12 +b^22 +b^23 )

Since oa=i+ 2 j− 3 k,

a 1 = 1 ,a 2 =2anda 3 =− 3

Since ob= 2 i−j+ 4 k,

b 1 = 2 ,b 2 =−1andb 3 = 4

Thus,

cosθ=

( 1 × 2 )+( 2 ×− 1 )+(− 3 × 4 )

√

( 12 + 22 +(− 3 )^2 )

√

( 22 +(− 1 )^2 + 42 )

=

− 12

√

14

√

21

=− 0. 6999

i.e.θ= 134. 4 ◦or 225. 6 ◦.

By sketchingthe positionof the two vectors as shown in

Problem1,it will beseenthat 225.6◦is not an acceptable

answer.

Thus the angle between the vectors oa and ob,

θ=134.4◦

Direction cosines

From Fig. 26.2, or=xi+yj+zk and from

equation (3),|or|=

√

x^2 +y^2 +z^2.

Iformakes angles ofα,βandγwith the co-ordinate

axesi,jandkrespectively, then:

The direction cosines are:

cosα=

x

√

x^2 +y^2 +z^2

cosβ=

y

√

x^2 +y^2 +z^2

and cosγ=

y

√

x^2 +y^2 +z^2

such that cos^2 α+cos^2 β+cos^2 γ=1.

The values of cosα,cosβand cosγare called the

direction cosinesofor.

Problem 5. Find the direction cosines of

3 i+ 2 j+k.

√

x^2 +y^2 +z^2 =

√

32 + 22 + 12 =

√

14