Mathematical Methods for Physics and Engineering - Matematica.NET

7.6 MULTIPLICATION OF VECTORSIf we introduce a set of basis vectors that are mutually orthogonal, such as i, j,k, we can write the components of a vector a, with respect to that basis, in termsof the scalar product of a with each of the basis vectors, i.e. a x = a·i, a y = a·j anda z = a · k. In terms of the components a x , a y and a z the scalar product is given bya · b =(a x i + a y j + a z k) · (b x i + b y j + b z k)=a x b x + a y b y + a z b z , (7.21)where the cross terms such as a x i · b y j are zero because the basis vectors aremutually perpendicular; see equation (7.18). It should be clear from (7.15) thatthe value of a · b has a geometrical definition and that this value is independentof the actual basis vectors used.◮Find the angle between the vectors a = i +2j +3k and b =2i +3j +4k.From (7.15) the cosine of the angle θ between a and b is given bycos θ = a · b|a||b| .From (7.21) the scalar product a · b has the valuea · b =1× 2+2× 3+3× 4=20,and from (7.13) the lengths of the vectors areThus,|a| = √ 1 2 +2 2 +3 2 = √ 14 and |b| = √ 2 2 +3 2 +4 2 = √ 29.cos θ =20√14√29≈ 0.9926 ⇒ θ =0.12 rad. ◭We can see from the expressions (7.15) and (7.21) for the scalar product that ifθ is the angle between a and b thencos θ = a x b xa b + a y b ya b + a z b za bwhere a x /a, a y /a and a z /a are called the direction cosines of a, since they give thecosine of the angle made by a with each of the basis vectors. Similarly b x /b, b y /band b z /b are the direction cosines of b.If we take the scalar product of any vector a with itself then clearly θ = 0 andfrom (7.15) we havea · a = |a| 2 .Thus the magnitude of a can be written in a coordinate-independent form as|a| = √ a · a.Finally, we note that the scalar product may be extended to vectors withcomplex components if it is redefined asa · b = a ∗ xb x + a ∗ yb y + a ∗ zb z ,where the asterisk represents the operation of complex conjugation. To accom-221