Scalar and vector products 277
Let a = a 1 i + a 2 j + a 3 k
and b = b 1 i + b 2 j + b 3 k
a • b = (a 1 i + a 2 j + a 3 k) • (b 1 i + b 2 j + b 3 k)
Multiplying out the brackets gives:
a • b = a 1 b 1 i • i + a 1 b 2 i • j + a 1 b 3 i • k
+ a 2 b 1 j • i + a 2 b 2 j • j + a 2 b 3 j • k
+ a 3 b 1 k • i + a 3 b 2 k • j + a 3 b 3 k • k
However, the unit vectors i, j and k all have a magnitude
of 1 and i • i = (1)(1) cos 0 ◦ = 1, i • j = (1)(1) cos 90 ◦ = 0,
i • k = (1)(1) cos 90 ◦ = 0 and similarly j • j = 1, j • k = 0
and k • k = 1. Thus, only terms containing i • i, j • j or
k • k in the expansion above will not be zero.
Thus, the scalar product
a • b = a 1 b 1 + a 2 b 2 + a 3 b 3
(2)
Both a and b in equation (1) can be expressed in terms
of a 1 , b 1 , a 2 , b 2 , a 3 and b 3 .
c
P
b
a
A
B
O
Figure 26.7
From the geometry of Fig. 26.7, the length of diagonal
OP in terms of side lengths a, b and c can be obtained
from Pythagoras’ theorem as follows:
OP
2
= OB
2
+ BP
2 and
OB
2
= OA
2
+ AB
2
Thus, OP
2
= OA
2
+ AB
2
+ BP
2
= a
2
+ b
2
+ c
2
,
in terms of side lengths
Thus, the length or modulus or magnitude or norm of
vector OP is given by:
OP =
(a 2 + b
2
+ c 2 )
(3)
Relating this result to the two vectors a 1 i + a 2 j + a 3 k
and b 1 i + b 2 j + b 3 k, gives:
a =
(a 2
1 + a 2
2 + a 2
3 )
and b =
(b 2
1 + b 2
2 + b 2
3 ).
That is, from equation (1),
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 )
(4)
Problem 2. Find vector a joining points P and Q
where point P has co-ordinates (4, −1, 3) and point
Q has co-ordinates (2, 5, 0). Also, find |a|, the
magnitude or norm of a.
Let O be the origin, i.e. its co-ordinates are (0, 0, 0). The
position vector of P and Q are given by:
OP = 4i − j + 3k and OQ = 2i + 5j
By the addition law of vectors OP + PQ = OQ.
Hence a =PQ = OQ − OP
i.e.
a =PQ = (2i + 5j) − (4i − j + 3k)
= −2i + 6j − 3k
From equation (3), the magnitude or norm of a,
|a| =
(a 2 + b 2 + c 2 )
=
[(−2) 2 + 6 2 + (−3) 2 ] =
√
49 = 7
Problem 3. If p = 2i + j −k and q = i −3j + 2k
determine:
(i) p • q
(ii) p +q
(iii) |p + q|
(iv) |p| +|q|
Let a = a 1 i + a 2 j + a 3 k
and b = b 1 i + b 2 j + b 3 k
a • b = (a 1 i + a 2 j + a 3 k) • (b 1 i + b 2 j + b 3 k)
Multiplying out the brackets gives:
a • b = a 1 b 1 i • i + a 1 b 2 i • j + a 1 b 3 i • k
+ a 2 b 1 j • i + a 2 b 2 j • j + a 2 b 3 j • k
+ a 3 b 1 k • i + a 3 b 2 k • j + a 3 b 3 k • k
However, the unit vectors i, j and k all have a magnitude
of 1 and i • i = (1)(1) cos 0 ◦ = 1, i • j = (1)(1) cos 90 ◦ = 0,
i • k = (1)(1) cos 90 ◦ = 0 and similarly j • j = 1, j • k = 0
and k • k = 1. Thus, only terms containing i • i, j • j or
k • k in the expansion above will not be zero.
Thus, the scalar product
a • b = a 1 b 1 + a 2 b 2 + a 3 b 3
(2)
Both a and b in equation (1) can be expressed in terms
of a 1 , b 1 , a 2 , b 2 , a 3 and b 3 .
c
P
b
a
A
B
O
Figure 26.7
From the geometry of Fig. 26.7, the length of diagonal
OP in terms of side lengths a, b and c can be obtained
from Pythagoras’ theorem as follows:
OP
2
= OB
2
+ BP
2 and
OB
2
= OA
2
+ AB
2
Thus, OP
2
= OA
2
+ AB
2
+ BP
2
= a
2
+ b
2
+ c
2
,
in terms of side lengths
Thus, the length or modulus or magnitude or norm of
vector OP is given by:
OP =
(a 2 + b
2
+ c 2 )
(3)
Relating this result to the two vectors a 1 i + a 2 j + a 3 k
and b 1 i + b 2 j + b 3 k, gives:
a =
(a 2
1 + a 2
2 + a 2
3 )
and b =
(b 2
1 + b 2
2 + b 2
3 ).
That is, from equation (1),
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 )
(4)
Problem 2. Find vector a joining points P and Q
where point P has co-ordinates (4, −1, 3) and point
Q has co-ordinates (2, 5, 0). Also, find |a|, the
magnitude or norm of a.
Let O be the origin, i.e. its co-ordinates are (0, 0, 0). The
position vector of P and Q are given by:
OP = 4i − j + 3k and OQ = 2i + 5j
By the addition law of vectors OP + PQ = OQ.
Hence a =PQ = OQ − OP
i.e.
a =PQ = (2i + 5j) − (4i − j + 3k)
= −2i + 6j − 3k
From equation (3), the magnitude or norm of a,
|a| =
(a 2 + b 2 + c 2 )
=
[(−2) 2 + 6 2 + (−3) 2 ] =
√
49 = 7
Problem 3. If p = 2i + j −k and q = i −3j + 2k
determine:
(i) p • q
(ii) p +q
(iii) |p + q|
(iv) |p| +|q|