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 = 2i + j − k,
a 1 = 2, a 2 = 1 and a 3 =−1
and when q = i − 3j +2k,
b 1 = 1, b 2 = −3 and b 3 = 2
Hence p • q = (2)(1) + (1)(−3) + (−1)(2)
i.e.
p • q = −3
(ii) p +q = (2i + j −k) + (i − 3j +2k)
= 3i −2j + k
(iii) |p +q| =|3i − 2 j + k|
From equation (3),
|p + q| =
[3 2 + (−2) 2 + 1 2 ] =
√
14
(iv) From equation (3),
|p| = |2i + j − k|
=
[2 2 + 1 2 + (−1) 2 ] =
√
6
Similarly,
|q| = |i − 3 j + 2k|
=
[1 2 + (−3) 2 + 2 2 ] =
√
14
Hence |p| +|q|=
√
6 +
√
14 = 6.191, correct to 3
decimal places.
Problem 4. Determine the angle between vectors
oa and ob when
oa = i + 2j − 3k
and ob = 2i − j + 4k.
An equation for cos θ is given in equation (4)
cos θ =
a 1 b 1 + a 2 b 2 + a 3 b 3
(a 2
1 + a 2
2 + a 2
3 )
(b 2
1 + b 2
2 + b 2
3 )
Since oa = i + 2j − 3k,
a 1 = 1, a 2 = 2 and a 3 = −3
Since ob = 2i − j + 4k,
b 1 = 2, b 2 = −1 and b 3 = 4
Thus,
cos θ =
(1 × 2) + (2 × −1) + (−3 × 4)
(1 2 + 2 2 + (−3) 2 )
(2 2 + (−1) 2 + 4 2 )
=
−12
√
14
√
21
= −0.6999
i.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 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 .
If or makes angles of α, β and γ with the co-ordinate
axes i, j and k respectively, then:
The direction cosines are:
cos α =
x
x 2 + y 2 + z 2
cos β =
y
x 2 + y 2 + z 2
and cos γ =
y
x 2 + y 2 + z 2
such that cos 2 α + cos 2 β + cos 2 γ = 1.
The values of cos α, cos β and cos γ are called the
direction cosines of or.
Problem 5. Find the direction cosines of
3i + 2j +k.
x 2 + y 2 + z 2 =
3 2 + 2 2 + 1 2 =
√
14