Scalar and vector products 281
Squaring both sides of a vector product equation gives:
(|a × b|)
2
= a
2 b
2 sin
2
θ = a
2 b
2
(1 − cos
2
θ)
= a
2 b
2
− a
2 b
2 cos
2
θ
(6)
It is stated in Section 26.2 that a • b = ab cos θ, hence
a • a = a
2 cos θ.
But θ = 0
◦
, thus a • a = a
2
Also, cos θ =
a • b
ab
.
Multiplying both sides of this equation by a 2 b 2 and
squaring gives:
a
2 b
2 cos
2
θ =
a 2 b 2 (a • b) 2
a 2 b 2
= (a • b)
2
Substituting in equation (6) above for a 2 = a • a, b 2 = b • b
and a
2 b
2 cos
2
θ = (a • b)
2 gives:
(|a × b|)
2
= (a • a)(b • b) − (a • b)
2
That is,
|a × b| =
[(a • a)(b • b) − (a • b)
2 ]
(7)
Problem 7. For the vectors a =i + 4j −2k and
b =2i − j +3k find (i) a × b and (ii) |a × b|.
(i) From equation (5),
a × b =
i
j k
a 1 a 2 a 3
b 1 b 2 b 3
= i
a 2 a 3
b 2 b 3
− j
a 1 a 3
b 1 b 3
+ k
a 1 a 2
b 1 b 2
Hence
a × b =
i
j
k
1
4 −2
2 −1
3
= i
4 −2
−1
3
− j
1 −2
2
3
+ k
1
4
2 −1
= i(12 − 2) − j(3 + 4) + k(−1 − 8)
= 10i − 7j −9k
(ii) From equation (7)
|a × b| =
[(a • a)(b • b) − (a • b) 2 ]
Now
a • a = (1)(1) + (4 × 4) + (−2)(−2)
= 21
b • b = (2)(2) + (−1)(−1) + (3)(3)
= 14
and
a • b = (1)(2) + (4)(−1) + (−2)(3)
= −8
Thus |a × b| =
(21 × 14 − 64)
=
√
230 = 15.17
Problem 8. If p = 4i + j −2k, q =3i − 2j + k and
r = i −2k find (a) ( p −2q) × r (b) p × (2r × 3q).
(a) ( p − 2q) × r = [4i + j − 2k
− 2(3i − 2j + k)] × (i − 2k)
= (−2i + 5j − 4k) × (i − 2k)
=
i j
k
−2 5 −4
1 0 −2
from equation (5)
= i
5 −4
0 −2
− j
−2 −4
1 −2
+ k
−2 5
1 0
= i(−10 − 0) − j(4 + 4)
+ k(0 − 5), i.e.
( p − 2q) × r = −10i − 8j −5k
(b) (2r × 3q) = (2i − 4k) × (9i − 6j + 3k)
=
i j k
2 0 −4
9 −6 3
= i(0 − 24) − j(6 + 36)
+ k(−12 − 0)
= −24i − 42j −12k