MODULE 5a and 5b (Stewart, Sections 12.2, 12.3) INTRO: In MATH 1114 vectors were written either as rows (a1, a2,..., an) or as columns a 1 a. ...

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1 MODULE 5a and 5b (Stewat, Sections 2.2, 2.3) INTRO: In MATH 4 vectos wee witten eithe as ows (a, a2,..., an) o as columns a a 2... a n and the set of all such vectos of fixed length n was called the vecto space R n. We shall be inteested only in the cases n = 2, 3, ie, only in R 2 and R 3. Editos (and we) pefe to have such vectos witten as ows in textbook explanations, since it takes up substantially less oom in displaying a paagaph, although if the vectos ae to be multiplied in the nomal way by a matix, then it is impotant to wite them as columns. Since we will not be using matices in this manne, we will geneally display ou vectos as ows. Vectos In R 2 o R 3 : < a,b > o < a, b, c > Some textbooks (including Stewat) distinguish between a vecto (in R 2 fo example) witten < a,b > and the point (a, b) in the plane R 2. This distinction is unimpotant, since any vecto < a,b > can equally be thought of as the aow going fom the oigin to the point (a, b). The angle which the aow (o a line segment) going fom the oigin to (a, b) makes with the positive x-axis (in R 2 ) is usually labeledθ. Moe on this angle shotly. Unit coodinate vectos in R 2 : î =,0,0, ĵ = 0,,0, ˆk = 0,0, (also witten as ˆx, ŷ, ẑ ). These point along the coodinate axes in the positive diection. Unit coodinate vectos in R 2 : î =,0, ĵ = 0, Moe geneal vecto space: The vecto spaces R n do not define what a vecto space is. In fact, they ae only special cases of the much moe geneal concept of a vecto space (although they ae especially impotant cases). In geneal a vecto space is any collection of objects with well-defined ules fo adding the objects and multiplying them by numbes in such a way that eight so-called vecto space axioms ae satisfied. (Two of these axioms (ules) ae, fo example, that u + v = v + u fo any of the objects u, v, and that k ( u + v ) = ku + kv fo any numbe k). When this is ealized, then the set of objects is called a vecto space and the objects themselves ae called the vectos. Of couse, R 2 and R 3 ae examples of sets of objects fo which thee ae well-defined ules fo adding and multiplying times numbes, and which satisfy the necessay vecto space axioms. Othe examples of vecto spaces, of cucial impotance fo many advanced topics in science and engineeing, might be the vecto space of all eal-valued functions defined on the inteval [0, ], the vecto space of all eal-valued continuous functions defined on the whole eal line, the vecto space of all

2 n n 2 matices, etc. (Indeed, we have no difficulty undestanding what is meant by adding two functions, o two n n 2 matices, o multiplying them times a numbe, and, as it tuns out, the eight axioms ae satisfied in these cases) In this couse, howeve, vecto always efes to the vecto spaces R 2 o R 3. Length (o nom) of a vecto. < a,b > = a 2 + b 2. < a, b, c > = a 2 + b 2 + c 2 Example: Length < 3, 2 >= < 3, 2 > = 3 Example: Length < 2,,-3 >= < 2,,-3 > = 4 Components of vectos If v =< a, b, c > (o v =< a,b >) then a, b, c (o a, b) ae called the components of v. Often we label components with a subscipt, usually x, y, z o, 2, 3, so if v =< 7, 3, 5 > then we may wite v x = 7, v y = 3, v z = 5, o we may wite v = 7, v 2 = 3, v 3 = 5. If v is a vecto in R 2 of length v, which makes an angle θ with the positive x-axis, then the components of v ae also given as v =< v cosθ, v sinθ > That is to say, if v is any vecto in two dimensions which is at an angle c as measued fom the positive x-axis, then v x = v cosθ and v y = v sinθ. This is a vey impotant fomula in applications, which can be veified by simple geomety. Note that thee is a simila, but moe complicated, fomula fo vectos in R 3, involving two angles, usually taken as θ along with the so-called azimuthal angle measued fom the z-axis, but we will not conside such poblems. Add, subtact, multiply by a constant: (thee ways) () by component. < 2, 3, 5 > + < 7,-2, > = < 9,, 6 >. < 2, 3, 5 > - < 7,-2, > = < -5, 5, 4 >, 5 < 2, 3, 5 > = < 0, 5, 25 >

3 (2) gaphically If a is a vecto dawn as an aow fom the oigin to the point a, then - a is the vecto coesponding to the aow dawn fom the oigin in exactly the opposite diection to that of a. The vecto k a if k is positive coesponds to the same aow as a but lengthened o shotened by a facto of k, and if k is negative coesponds to the same aow as - a but again lengthened o shotened by a facto of k. If a and b ae two vectos dawn as aows fom the oigin to the points a and b, then a + b is the aow dawn fom the oigin to the futhest vetex of the paallelogam with sides a and b. Anothe way to say this is that a + b may be obtained gaphically by combining a and b as aows tail to head. Note that a + b, the length of a + b, is just the length of the diagonal (which begins at the oigin) of the paallelogam. To obtain a - b gaphically, it is easiest to eplace b fist by - b and then add these as above. (3) by adding pojections (in the plane, say) ie, F + F 2 = F = F + F 2 F x = F cosθ + F 2 cosθ 2 F y = F sinθ + F 2 sinθ 2 F cosθ + F 2 cosθ 2, F sinθ + F 2 sinθ 2 whee the angles θ and θ 2 should be measued fom the positive x-axis to the vectos F and F 2. This is tue since the fist component of F, fo example, is just F cosθ, etc., so we ae just adding components.

4 Note: Unit vecto in diection of w is w w. This can be vey useful. Fo example, to find a vecto of length 3 in the diection of the vecto v, one can fist find the unit vecto in the diection of v and then multiply it by 3. The esult, of couse, would be 3 v v. Example: Find a point a distance 5 fom P = (, 2, 3) in the diection of u =< 2,, 2 >. Answe: Find a vecto of length 5 in the diection of u and add it to <, 2, 3 >. Thus: 5 <, 2, 3 > + u = <, 2, 3 > + 5 u 3 < 2,, 2 > = 3 3, 3, 9 so answe is the point 3 3 3, 3, 9 3. Note how we do not hesitate to switch between the notation fo a point (, 2, 3) and fo a vecto <, 2, 3 >. That is why some books do not even make a distinction in notation. Dot poduct < a,b > < c,d >= ac + bd < a, b, c > < d, e, f >= ad + be + cf Dot poduct gives length: v = v v. Check: < a, b, c > = a,b,c a,b,c = a 2 + b 2 + c 2 But dot poduct gives moe: Othogonality and Angle Two vectos v and w ae pependicula (othogonal) IFF v w = 0 Moe geneally: (LAW OF COSINES): so v w = cosθ = v w cosθ v w v w Example: Fo what k is <, 3 > othogonal to < 2, k >? (Solve 2 + 3k = 0) Example: Angle between < 2, > and < 3,-4 > cosθ = Example: Angle between < 2, > and positive x-axis. (Use <, 0 > as the second vecto)

5 NOTE: Because we daw all vectos fom the oigin, w will be PARALLEL to v only if w = k v fo some (positive o negative) constant k. (Vecto) Pojection of v onto w (Really onto diection of w ) Pojection v onto w v w = w 2 w Reason: If pojection of v onto w wee just a numbe, we would expect it to be v cos θ whee θ is the angle between v and w, i.e., the component of v in the w -diection, if w wee taken as a coodinate diection. In ou book, the pojection is an actual vecto, namely: v cosθ {unit vecto in diection of w } = v v w v w w w Example: Poj. of <, 2, 3 > onto < 2,, 2 > = 20 9, 0 9, 20 9 Note: Poj of < a,b > onto x-axis = Poj of < a,b > onto <, 0 > a,b,0. =,0 = a,0 = aî as it should be!

6 Statics If object is stationay, sum of foces on it must be zeo (one of Newton s Laws). It is usual to obtain simultaneous equations in statics poblems, setting the sum of components of all foces on object in the x-diection equal to zeo and the sum of components of all foces in the y-diection equal to zeo. Sometimes (inclined plane poblem, fo example) components along two pependicula diections not necessaily in the ˆx and ŷ diections ae set equal to zeo. Example: Object of weight 0 lbs. is suspended by cables at angles of 30 degees and 60 degees fom vetical. What is the tension (magnitude of the foce) on each cable? Answe: The foces on the object ae gavity, which is 0 lbs. in the downwad (- ŷ ) diection, and the two tensions, say T and T 2 pointing up the espective cables. So, assuming the 30 degee cable is to the ight, and noting that the 30 degee (fom vetical) cable actually has an angle of 60 degees with efeence to the positive x-axis and the 60 degee cable actually has an angle of 50 degees with efeence to the positive x-axis, we would need to solve the simultaneous equations: T cos π 3 + T 2 cos 5π 6 T sin π 3 + T 2 sin 5π 6 = 0 (foces in x-diection) 0 = 0 (foces in y-diection) fo the unknowns T and T2. QUIZ PREP. Opeations on vectos: adding, subtacting and multiplying by a constant. a) by components. (b) gaphically. (c) given in tems of lengths and angles 2. Length of a vecto 3. Locating points a given distance and diection away 4. Angle between vectos 5. Pojection of one vecto onto anothe