240 VECTOR GEOMETRY(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 =1 anda 3 =− 1
and whenq=i− 3 j+ 2 k,
b 1 =1,b 2 =−3 andb 3 = 2
Hence p•q=(2)(1)+(1)(−3)+(−1)(2)
i.e. p•q=− 3
(ii)p+q=(2i+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 ]=√
6Similarly,|q|=|i− 3 j+ 2 k|=√
[1^2 +(−3)^2 + 22 ]=√
14Hence|p|+|q|=√
6 +√
14 =6.191, correct
to 3 decimal places.Problem 4. Determine the angle between vec-
torsoaandobwhenoa=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^21 +a^22 +a^23 )√
(b^21 +b^22 +b^23 )Since oa=i+ 2 j− 3 k,
a 1 =1,a 2 =2 anda 3 =− 3Since ob= 2 i−j+ 4 k,
b 1 =2,b 2 =−1 andb 3 = 4Thus,cosθ=(1×2)+(2×−1)+(− 3 ×4)
√
(1^2 + 22 +(−3)^2 )√
(2^2 +(−1)^2 + 42 )=− 12
√
14√
21=− 0. 6999i.e.θ= 134. 4 ◦or 225. 6 ◦.By sketching the position of the two vectors as
shown in Problem 1, it will be seen that 225.6◦is
not an acceptable answer.
Thus the angle between the vectorsoaandob,
θ=134.4◦.Direction cosinesFrom Fig. 22.2,or=xi+yj+zkand from equa-
tion (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^2cosβ=y
√
x^2 +y^2 +z^2and cosγ=y
√
x^2 +y^2 +z^2such 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 =√
14The direction cosines are:cosα=x
√
x^2 +y^2 +z^2=3
√
14=0.802cosβ=y
√
x^2 +y^2 +z^2=2
√
14=0.535and cosγ=y
√
x^2 +y^2 +z^2=1
√
14=0.267(and hence α=cos−^10. 802 = 36. 7 ◦, β=cos−^1
0. 535 =57.7◦andγ=cos−^10. 267 = 74. 5 ◦).