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