19.3: Appendix - Vector algebra
( \newcommand{\kernel}{\mathrm{null}\,}\)
Linear operations
The important force fields in classical mechanics, namely, gravitation, electric, and magnetic, are vector fields that have a position-dependent magnitude and direction. Thus, it is useful to summarize the algebra of vector fields.
A vector a has both a magnitude |a| and a direction defined by the unit vector ˆea, that is, the vector can be written as a bold character a where
a=a⋅ˆea
where by convention the implied modulus sign is omitted. The hat symbol on the vector ˆea designates that this is a unit vector with modulus |ˆea|=1.
Vector force fields are assumed to be linear, and consequently they obey the principle of superposition, are commutative, associative, and distributive as illustrated below for three vectors a,b,c plus a scalar multiplier γ.
a±b=±b+aa+(b+c)=(a+b)+cγ(a+b)=γa+γb
The manipulation of vectors is greatly facilitated by use of components along an orthogonal coordinate system defined by three orthogonal unit vectors (ˆe1,ˆe2,ˆe3). For example the cartesian coordinate system is defined by three unit vectors which, by convention, are called (ˆi,ˆj,ˆk).
Scalar product
Multiplication of two vectors can produce a 9−component tensor that can be represented by a 3×3 matrix as discussed in appendix 19.5. There are two special cases for vector multiplication that are important for vector algebra; the first is the scalar product, and the second is the vector product.
The scalar product of two vectors is defined to be
a⋅b=|a||b|cosθ
where θ is the angle between the two vectors. It is a scalar and thus is independent of the orientation of the coordinate axis system. Note that the scalar product commutes, is distributive, and associative with a scalar multiplier, that is
a⋅b=b⋅aa⋅(b+c)=a⋅b+a⋅c(λa)⋅b=λ(b⋅a)
Note that a⋅a=|a|2 and if a and b are perpendicular then cosθ=0 and thus a⋅b=0
If the three unit vectors (ˆe1,ˆe2,ˆe3) form an orthonormal basis, that is, they are orthogonal unit vectors, then from equations ??? and ???
ˆei⋅ˆek=δik
If ˆa is the unit vector for the vector a then the scalar product of a vector a with one of these unit vectors ˆen gives the cosine of the angle between the vector a and ˆen, that is
a⋅ˆe1=|a|(ˆa⋅ˆe1)=|a|cosαa⋅ˆe2=|a|(ˆa⋅ˆe2)=|a|cosβa⋅ˆe3=|a|(ˆa⋅ˆe3)=|a|cosγ
where the cosines are called the direction cosines since they define the direction of the vector a with respect to each orthogonal basis unit vector. Moreover, a⋅ˆe1=|a|ˆa⋅ˆe1=|a|cosα is the component of a along the ˆe1 axis. Thus the three components of the vector a is fully defined by the magnitude |a| and the direction cosines, corresponding to the angles α,β,γ. That is,
a1=|a|(ˆa⋅ˆe1)=|a|cosαa2=|a|(ˆa⋅ˆe2)=|a|cosβa3=|a|(ˆa⋅ˆe3)=|a|cosγ
If the three unit vectors (ˆe1,ˆe2,ˆe3) form an orthonormal basis then the vector is fully defined by
a=a1ˆe1+a2ˆe2+a3ˆe3
Consider two vectors
a=a1ˆe1+a2ˆe2+a3ˆe3
b=b1ˆe1+b2ˆe2+b3ˆe3
Then using ???
a⋅b=a1b1+a2b2+a3b3=|a||b|cosθ
where θ is the angle between the two vectors. In particular, since the direction cosine cosαa=a1|a|, then Equation ??? gives
cosθ=cosαacosαb+cosβacosβb+cosγacosγb
Note that when θ=0 then ??? gives
cos2α+cos2β+cos2γ=1
Vector product
The vector product of two vectors is defined to be
c=a×b=|a||b|sinθˆn
where θ is the angle between the vectors and ˆn is a unit vector perpendicular to the plane defined by a and b such that the unit vectors (ˆa,ˆb,ˆn) obey a right-handed screw rule. The vector product acts like a pseudovector which comprises a normal vector multiplied by a sign factor that depends on the handedness of the system as described in appendix 19.4.3.
The components of c are defined by the relation
ci≡∑jkεijkajbk
where the (Levi-Civita) permutation symbol εijk has the following properties
εijk=0 if an index is equal to any another indexεijk=+1 if i,j,k, form an even permutation of 1,2,3εijk=−1 if i,j,k, form an odd permutation of 1,2,3
For example, if the three unit vectors (ˆe1,ˆe2,ˆe3) form an orthonormal basis, then ˆei≡∑jkεijkˆejˆek, i.e.
ˆe1׈e2=ˆe3ˆe2׈e3=ˆe1ˆe3׈e1=ˆe2ˆe2׈e1=−ˆe3ˆe3׈e2=−ˆe1ˆe1׈e3=−ˆe2ˆe1׈e1=0ˆe2׈e2=0ˆe3׈e0=0
The vector product anticommutes in that
a×b=−b×a
However, it is distributive and associative with a scalar multiplier
a×(b+c)=a×b+a×c
(λa)×b=λ(a×b)
Note that when sinθ=0 then a×b=0 and in particular, a×a=0.
Consider two vectors
a=a1ˆe1+a2ˆe2+a3ˆe3
b=b1ˆe1+b2ˆe2+b3ˆe3
Then using equations ??? and 19.3.21 − 19.3.23
a×b=|a||b|sinθ=|ˆe1ˆe2ˆe3a1a2a3b1b2b3|=ˆe1(a2b3−a3b2)+ˆe2(a3b1−a1b3)+ˆe3(a1b2−a2b1)
where θ is the angle between the two vectors and the determinant is evaluated for the top row. Examples of vector products are torque N=r×F, angular momentum L=r×p, and the magnetic force FB=qv×B.
Triple products
The following scalar and vector triple products can be formed from the product of three vectors and are used frequently.
Scalar triple products
There are several permutations of scalar triple products of three vectors [a,b,c] that are identical.
a⋅(b×c)=c⋅(a×b)=b⋅(c×a)=(a×b)⋅c=−a⋅(c×b)
That is, the scalar product is invariant to cyclic permutations of the three vectors but changes sign for interchange of two vectors. The scalar product is unchanged by swapping the scalar (dot) and vector (cross).
Because of the symmetry the scalar triple product can be denoted as [a,b,c] and
[a,b,c]>0 if [a,b,c] is right-handed[a,b,c]=0 if [a,b,c] is coplanar[a,b,c]<0 if [a,b,c] is left-handed
The scalar triple product can be written in terms of the components using a determinant
[a,b,c]=|a1a2a3b1b2b3c1c2c3|
Vector triple product
The vector triple product a×(b×c) is a vector. Since (b×c) is perpendicular to the plane of b,c, then a×(b×c) must lie in the plane containing b,c. Therefore the triple product can be expanded in terms of b,c, as given by the following identity
a×(b×c)=(a⋅c)b−(a⋅b)c
Problems
1. Partition the following exercises among your collaborators. Once you have completed your problem, check with a classmate before writing it on the board. After you have verified that you have found the correct solution, write your answer in the space provided on the board, taking care to include the steps that you used to arrive at your solution. The following information is needed.
a=3i+2j−9k | b=−2i+3k | c=−2i+j−6k | d=i+9j+4k |
E=(27−431−2−205) | F=(3456) | G=(2−471−11) | H=(−8−1−3−42−2−100) |
Calculate each of the following
1. |a−(b+3c)| | 7. (EH)T |
2. Component of c along a | 8. |HE| |
3. Angle between c and d | 9. EHG |
4. (b×d)⋅a | 10. EG−HG |
5. (b×d)×a | 11. EH−HTET |
6. b×(d×a) | 12. F−1 |
2. For what values of a are the vectors A=2aˆi−2ˆj+aˆk and B=aˆi+2aˆj+2ˆk perpendicular?
3. Show that the triple scalar product (A×B)⋅C can be written as
(A×B)⋅C=|A1A2A3B1B2B3C1C2C3|
Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of A,B,C as long as they are in cyclic order, that is
(A×B)⋅C=A⋅(B×C)=B⋅(C×A)=(C×A)⋅B
Therefore we may use the notation ABC to denote the triple scalar product. Finally give a geometric interpretation of ABC by computing the volume of the parallelepiped defined by the three vectors A,B,C.