Higher Engineering Mathematics

SCALAR AND VECTOR PRODUCTS 243

D

Also, cosθ=

a•b

ab

.

Multiplying both sides of this equation bya^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 fora^2 =a•a,

b^2 =b•banda^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 vectorsa=i+ 4 j− 2 kand

b= 2 i−j+ 3 kfind (i)a×band (ii)|a×b|.

(i) From equation (5),

a×b=

∣

∣

∣

∣

∣

ijk

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=

∣

∣

∣

∣

∣

ijk

14 − 2

2 − 13

∣

∣

∣

∣

∣

=i

∣

∣

∣

∣

4 − 2

− 13

∣

∣

∣

∣−j

∣

∣

∣

∣

1 − 2

23

∣

∣

∣

∣

+k

∣

∣

∣

∣

14

2 − 1

∣

∣

∣

∣

=i(12−2)−j(3+4)+k(− 1 −8)

= 10 i− 7 j− 9 k

(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. Ifp= 4 i+j− 2 k,q= 3 i− 2 j+k

andr=i− 2 kfind (a) (p− 2 q)×r

(b)p×(2r× 3 q).

(a) (p− 2 q)×r=[4i+j− 2 k

−2(3i− 2 j+k)]×(i− 2 k)

=(− 2 i+ 5 j− 4 k)×(i− 2 k)

=

∣ ∣ ∣ ∣ ∣ ∣

ij k

− 25 − 4

10 − 2

∣ ∣ ∣ ∣ ∣ ∣

from equation (5)

=i

∣

∣

∣

∣

5 − 4

0 − 2

∣

∣

∣

∣−j

∣

∣

∣

∣

− 2 − 4

1 − 2

∣

∣

∣

∣

+k

∣

∣

∣

∣

− 25

10

∣

∣

∣

∣

=i(− 10 −0)−j(4+4)

+k(0−5), i.e.

(p− 2 q)×r=− 10 i− 8 j− 5 k

(b) (2r× 3 q)=(2i− 4 k)×(9i− 6 j+ 3 k)

=

∣ ∣ ∣ ∣ ∣ ∣

ijk

20 − 4

9 − 63

∣ ∣ ∣ ∣ ∣ ∣

=i(0−24)−j(6+36)

+k(− 12 −0)

=− 24 i− 42 j− 12 k

Hence

p×(2r× 3 q)=(4i+j− 2 k)

×(− 24 i− 42 j− 12 k)

=

∣ ∣ ∣ ∣ ∣ ∣

ijk

41 − 2

− 24 − 42 − 12

∣ ∣ ∣ ∣ ∣ ∣