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