Higher Engineering Mathematics, Sixth Edition

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 thata•b=abcosθ, hence

a•a=a^2 cosθ.

Butθ= 0 ◦,thusa•a=a^2

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

Substitutinginequation(6)abovefora^2 =a•a,b^2 =b•b
anda^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+kand

r=i− 2 kfind (a)(p− 2 q)×r (b)p×( 2 r× 3 q).

(a)(p− 2 q)×r=[4i+j− 2 k

− 2 ( 3 i− 2 j+k)]×(i− 2 k)

=(− 2 i+ 5 j− 4 k)×(i− 2 k)

=

∣ ∣ ∣ ∣ ∣ ∣ ∣

ij k

− 25 − 4

10 − 2

∣ ∣ ∣ ∣ ∣ ∣ ∣

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

=

∣ ∣ ∣ ∣ ∣ ∣ ∣

ijk

20 − 4

9 − 63

∣ ∣ ∣ ∣ ∣ ∣ ∣

=i( 0 − 24 )−j( 6 + 36 )

+k(− 12 − 0 )

=− 24 i− 42 j− 12 k