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Vector Analysis: MS4613 Vector Analysis: MS4613 Prof. S.B.G. O’Brien Main Building, B3041. References Advanced engineering mathematics, Kreyszig. Mathematics applied to continuum mechanics, Segel. Mathematical methods for physicists, Arfken. Mathematics applied to the deterministic sciences, Lin and Segel. Vectors and cartesian tensors, Bourne and Kendall. Vector Analysis (Schaum series), Murray and Spiegel. GREEK ALPHABET For each Greek letter, we illustrate the form of the capital letter and the form of the lower case letter. In some cases, there is a popular variation of the lower case letter. Greek Greek English Greek Greek English letter name equivalent letter name equivalent A α Alpha a N ν Nu n B β Beta b Ξ ξ Xi x Γ γ Gamma g O o Omicron o ∆ δ Delta d Π π ̟ Pi p E ǫ ε Epsilon e P ρ ̺ Rho r Z ζ Zeta z Σ σ ς Sigma s H η Eta e T τ Tau t Θ θ ϑ Theta th Υ υ Upsilon u I ι Iota i Φ φ ϕ Phi ph K κ Kappa k X χ Chi ch Λ λ Lambda l Ψ φ Psi ps M µ Mu m Ω ω Omega o Exam Papers: http://www.ul.ie/portal/students Some papers are also available at: http://www.staff.ul.ie/obriens/index.html 1 1 Rectangular cartesian coordinates and rotation of axes 1.1 Rectangular cartesian coordinates We wish to be able to describe locations and directions in space. From a fixed point O, called the origin (of coordinates), draw three fixed lines Ox,Oy,Oz at right angles to each other as in fig.1.1 and forming a right handed system (rectangular cartesian axes) (use RH screw rule as in fig.1.2). If we are restricting attention to a 2D flat surface (e.g. tabletop) then two axes are sufficient (usually the x and y axes). Any point P in space can be uniquely represented as a triad (x,y,z) (or pair(x,y) in the 2D case) of real numbers called the coordinates of P. Thus we write P(x,y,z) or P(x,y). N.B. Don’t confuse the distance x with the ‘x axis’. 1.1.1 Distance from the origin To find the distance of a point P from the origin, construct a parallelopiped which has PL,PM,PN as edges (fig.1.3). From Pythagoras’ theorem we have: OP 2 = ON 2 + PN 2 = PL2 + PM 2 + PN 2 Since the perpendicular distances of p from the coordinate planes are x , y , z it follows that: | | | | | | OP = (x2 + y2 + z2) (1) p 1.1.2 Distance between points The distance between the points P with coordinates (x,y,z) relative to Ox,Oy,Oz and P ′ with coordi- nates (x′,y′,z′) is found in the following way. Through P construct three new coordinate axes PX,PY,PZ parallel to the original axes Ox,Oy,Oz. Let the coordinates of P ′ relative to the new axes be X,Y,Z. Then it is easily seen that: ′ ′ ′ x + X = x , y + Y = y , z + Z = z and it follows that ′ ′ ′ X = x x, Y = y y,Z = z z. − − − Now using (1.1) we find: ′ P P = (X2 + Y 2 + Z2) p and so in terms of coordinates relative to the original axes: ′ P P = (x′ x)2 + (y′ y)2 + (z′ z)2 (2) p − − − (See Fig. 1.4). 1.2 Direction cosines and direction ratios 1.2.1 Direction cosines Definition Consider an arbitrary line OP. If the angles which OP makes with Ox,Oy,Oz (fig. 1.5) are α,β,γ then the direction cosines are defined to be cos α, cos β, cos γ and for convenience we write: l = cos α, m = cos β,n = cos γ (3) The direction cosines of the x-axis, for example, are 1,0,0. (The direction cosines only give information about the direction of the line but say nothing about its length). Note also that when working in 2D xy space, we have γ = π/2 and so n = 0 and we only usually speak of two direction cosines l,m. Denote the foot of the perpendicular from P to the x-axis by N, let OP = r and suppose that the coordinates of P are (x,y,z). From the triangle OPN we have ON = x = r cos α. Also if α is an acute angle, cos α and x are positive, while if α is obtuse, then they are| both| negative. It follows that x = r cos α and similarly y = r cos β,z = r cos γ. The direction cosines of OP are thus l = x/r, m = y/r, n = z/r. (4) Since r2 = x2 + y2 + z2, we have x2 + y2 + z2 r2 l2 + m2 + n2 = = = 1 (5) r2 r2 2 Thus the direction cosines of a line are not independent: they must satisfy (5) or they cannot be direction cosines of any line. The direction cosines of any line not passing through the origin are defined to be the same as those of the parallel line drawn from the origin in the same sense. (This again emphasizes the fact that direction cosines define a direction in space: no more and no less. We will see presently that a unit vector does the same thing in effect). 1.3 Angles between lines through the origin Given the direction cosines of two lines (which contain all information about the directions of the lines), we would like to find an expression for the angle between the lines. Consider two lines OA and OA′ with direction cosines l,m,n and l′,m′,n′. To find the angle θ between them, denote by B,B′ the points on OA, OA′ (produced if necessary) such that OB = OB′ = 1. (Fig.1.6). Then the coordinates of B,B′ are (l,m,n), (l′,m′,n′), using (4) with r=1. Applying the cosine rule to the triangle OBB′ gives 2 ′2 ′2 OB + OB BB 1 ′ cosθ = − =1 BB 2 2OB.OB′ − 2 as OB = OB′ =1. But from ( 2 ) ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ BB 2 = (l l)2 + (m m)2 + (n n)2 = (l 2 + m 2 + n 2) + (l2 + m2 + n2) 2(ll + mm + nn ) − − − − Using the results l2 + m2 + n2 =1,l′2 + m′2 + n′2 = 1, we obtain ′ ′ ′ cosθ = ll + mm + nn (6) Note that because cos(2π θ) = cosθ, we can still obtain (6) when the angle between OA and OA′ is taken as 2π θ. − − 1.3.1 Condition for perpendicular lines Two lines through the origin are perpendicular if and only if ′ ′ ′ ll + mm + nn =0 (7) Proof ( ). Two lines OA, OA′ are at right angles θ = π/2 or θ =3π/2 cos θ = 0. The result follows from (6).⇒ ⇒ ⇒ ( ). ll′ + mm′ + nn′ =0 cos θ =0 θ = π/2 or 3π/2 the two lines are perpendicular . ⇐ ⇒ ⇒ ⇒ | Example Find the angle between the lines OA and OB where A = (1, 3, 1) and B = (1, 2, 3). Using (4), we note that for A, r = √1+9+1= √11 while for B, r = √1+4+9= √14. So the direction cosines of A are 1/√11, 3/√11, 1/√11 and the direction cosines for B are 1/√14, 2/√14, 3/√14. Applying (6) we thus have: cos θ =(1+6+3)/√154 = θ = arccos((1 + 6 + 3)/√154) . ⇒ | If and only if “If and only if” as it occurred in the proof above is very often written as . It corresponds to in both directions. “if” corresponds to and “only if” corresponds to . For⇔ example, a if b is equivalent⇒ to a b or b a. ⇐ ⇒ Suppose “a,b”⇐ are two⇒ (mathematical) statements. Thus a b means that “if a is true then b is true and if b is true then a is true” i.e., it is equivalent to a b AND⇔ b a. In this context the terminology “necessary, sufficient” is also used. a b can be written⇒ “a is a sufficient⇒ condition for b to be true” or “b is a necessary condition for a to⇒ be true”. a b can be written “ b is a necessary and sufficient condition for a to be true and vice versa”. ⇔ 3 Contrapositive Note that if a b we can read this as “ b is a necessary condition for a to be true”. Thus if b is not true, we can infer⇒ that a is not true. This is referred to as the “contrapositive” of the statement and is commonly used in mathematical proofs. The original statement and the contrapositive are logically equivalent. If we write: a b, the contrapositive is not(b) not(a) where not(b) means b is not true. The only way that a b can⇒ fail to be correct is if the hypothesis⇒ a is true and b is false. Otherwise it is correct. Similarly, the⇒ only way in which the statement not(b) not(a) can fail to be true is if the hypothesis “not (b)” is true and the conclusion “not (a)” is false. This⇒ is the same as saying that b is false and a is true. And this, in turn, is precisely the situation in which a b fails to be correct. The two statements are either both correct or incorrect: they are logically equivalent.⇒ Thus we can accept a proof of the statement not(b) not(a) as a proof of the statement a b. ⇒ ⇒ Converse Another statement related to a b is b a.
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