When vector oa is multiplied by a scalar quantity, say k, the magnitude of the resultant vector will be k times the magnitude of oa and its direction will remain the same. Thus 2×(5 N at 20◦) results in a vector of magnitude 10N at 20◦. One of the products of two vector quantities is called the scalar or dot product of two vectors and is defined as the product of their magnitudes multiplied by the cosine of the angle between them. The scalar product of oa and ob is shown as oa • ob. For vectors oa=oa at θ1, and ob=ob at θ2 where θ2 >θ1, the scalar product is:

oa •ob = oa ob cos(θ2 − θ1) For vectors v1 and v2 shown in Fig. 28.4, the scalar product is:

v1 •v2 = v1v2 cosθ

The commutative law of algebra, a ×b=b×a applies to scalar products. This is demonstrated in Fig. 28.5. Let oa represent vector v1 and ob represent vector v2. Then:

oa •ob = v1v2 cos θ (by definition of a scalar product)

Similarly, ob • oa=v2v1 cosθ =v1v2 cosθ by the commutative law of algebra. Thus oa • ob=ob • oa

The projection of ob on oa is shown in Fig. 28.6(a) and by the geometry of triangle obc, it can be seen that the projection is v2 cosθ . Since, by definition

oa •ob = v1(v2 cosθ), it follows that

oa •ob = v1 (the projection of v2 on v1)

Similarly the projection of oa on ob is shown in Fig. 28.6(b) and is v1 cosθ . Since by definition

ob •oa = v2(v1 cosθ), it follows that

ob •oa = v2(the projection of v1 on v2) This shows that the scalar product of two vectors is the product of the magnitude of one vector and the magnitude of the projection of the other vector on it. The angle between two vectors can be expressed in terms of the vector constants as follows: Because a •b=a b cosθ ,

then cosθ = a • b ab (1)

Let a = a1i + a2 j + a3k and b = b1i + b2 j + b3k

a •b = (a1i + a2 j + a3k) • (b1i + b2 j + b3k)

Multiplying out the brackets gives:

a •b = a1b1i • i + a1b2i • j + a1b3i •k + a2b1 j • i + a2b2 j • j + a2b3 j •k + a3b1k • i + a3b2k • j + a3b3k •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 = a1b1 + a2b2 + a3b3 (2)

Both a and b in equation (1) can be expressed in terms of a1, b1, a2, b2, a3 and b3

From the geometry of Fig. 28.7, the length of diagonal OP in terms of side lengths a, b and c can be obtained from Pythagoras’ theorem as follows:

OP2 = OB2 + BP2 and OB2 = OA2 + AB2

Thus, OP2 = OA2 + AB2 + BP2

= a2 + b2 + c2, in terms of side lengths

Thus, the length or modulus or magnitude or norm of vector OP is given by:

OP = √

(a2 + b2 + c2) (3) Relating this result to the two vectors a1i+a2 j+ a3k and b1i+b2 j+b3k, gives:

a = √

and b = √

That is, from equation (1),

cos θ = a1b1 + a2b2 + a3b3√ (a21 + a22 + a23)

√ (b21 + b22 + b23)

(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| = √

(a2 + b2 + c2)

= √

[(−2)2 + 62 + (−3)2] = √

49 = 7

Problem 3. If p=2i+ j−k and q= i−3j+2k determine: (a) p •q (b) p+q (c) |p+q| (d) |p|+ |q|

(a) From equation (2),

if p = a1i + a2 j + a3k and q = b1i + b2 j + b3k then p •q = a1b1 + a2b2 + a3b3 When p = 2i+ j−k,

a1 = 2, a2 = 1 and a3 =−1 and when q = i−3j+2k,

b1 = 1, b2 = −3 and b3 = 2 Hence p •q = (2)(1)+ (1)(−3)+ (−1)(2) i.e. p •q = −3

(b) p+q=(2i+ j−k)+(i−3j+2k) =3i−2j+k

(c) |p+q|=|3i −2 j + k| From equation (3),

|p+q| = √

[32 + (−2)2 + 12] = √

(d) From equation (3),

|p| = |2i + j − k| = √

[22 + 12 + (−1)2] = √

Similarly,

|q| = |i − 3 j + 2k| = √

[12 + (−3)2 + 22] = √

Hence |p|+ |q|=√6+√14=6.191, correct to 3 decimal places.