Vectors , (0,0). 5. A vector is commonly denoted by putting an arrow above its symbol, as in the picture above. Here are some 3dimensional vectors:


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1 Vectors I ll look t vectors from n lgeric point of view nd geometric point of view. Algericlly, vector is n ordered list of (usully) rel numers. Here re some 2dimensionl vectors: (2, 3), ( ) 2, 3, (0,0). 5 The numers which mke up the vector re the vector s components. Here re some 3dimensionl vectors: ( ) 1 (1,2, 17), 2, 17,π, (0,0,0). Since we usully use x, y, nd z s the coordinte vriles in 3 dimensions, vector s components re sometimes referred to s its x, y, nd zcomponents. For instnce, the vector (1, 2, 17) hs xcomponent 1, it hs ycomponent 2, nd it hs zcomponent 17. The set of 2dimensionl relnumer vectors is denoted R 2, just like the set of ordered pirs of rel numers. Likewise, the set of 3dimensionl relnumer vectors is denoted R 3. Geometriclly, vector is represented y n rrow. Here re some 2dimensionl vectors: x A vector is commonly denoted y putting n rrow ove its symol, s in the picture ove. Here re some 3dimensionl vectors: z x y The reltionship etween the lgeric nd geometric descriptions comes from the following fct: The vector from point P(,) to point Q(c,d) is given y PQ = (c,d ). In 3 dimensions, the vector from point P(,,c) to point Q(d,e,f) is PQ = (d,e,f c). Remrk. You ve proly lredy noticed the following hrmless confusion: (3, 2) cn denote the point (3,2) in the xyplne, or the 2dimensionl rel vector (3,2). Notice tht the vector from the origin (0,0) 1
2 to the point (3,2) is the vector (3,2). 2 (3,2) (0,0) 3 So we cn usully regrd them s interchngele. When there s need to mke distinction, I will cll it out. Exmple. () Find the vector from P(3,4, 7) to Q( 2,2,5). () Find the vectors AB, BA, AC, nd CD for the points A(1,1), B(2,3), C( 2,0), nd D( 1,2). Sketch the vectors AB nd CD. () () PQ = ( 2 3,2 4,5 ( 7)) = ( 5, 2,12). AB = (2 1,3 1) = (1,2), BA = (1 2,1 3) = ( 1, 2), AC = ( 2 1,0 1) = ( 3, 1), CD = ( 1 ( 2),2 0) = (1,2). Notice tht BA = AB; this is true in generl. Here s sketch of the vectors AB nd CD: y B D A C x AB nd CD re oth (1,2); in the picture, you cn see tht the rrows which represent the vectors hve the sme length nd the sme direction. Geometriclly, two vectors (thought of s rrows) re equl if they hve the sme length nd point in the sme direction. 2
3 Exmple. In the picture elow, ssume the two lines re prllel. Which of the vectors, c, d is equl to the vector? is not equl to ; it hs the sme direction, ut not the sme length. c is not equl to ; it hs the sme length, ut the opposite direction. d is equl to, since it hs the sme length nd direction. Algericlly, two vectors re equl if their corresponding components re equl. Exmple. Find nd such tht (+2, ) = ( 8,7). Set the corresponding components equl nd solve for nd : + 2 = 8 = 7 3 = 15 / 3 3 = 5 Sustituting this into +2 = 8, I get 10 = 8, so = 2. The solution is = 2, = 5. The length of geometric vector is the length of the rrow tht represents it. The length of n lgeric vector is given y the distnce formul. If v = (,,c), the length of v is A vector with length 1 is clled unit vector. v = c 2. Exmple. () Find the length of (3,12, 4). ( 4 () Show tht 5, 3 ) is unit vector. 5 () (3,12, 4) = ( 4) 2 = 13. 3
4 () ( 4 5, 3 ) ( 4 ) 2 = ( ) = = 1. Algericlly, you dd or sutrct vectors y dding or sutrcting corresponding components: (,)+(c,d) = (+c,+d), (,) (c,d) = ( c, d). (Use n nlogous procedure to dd or sutrct 3dimensionl vectors.) You cn t dd or sutrct vectors with different numers of components. For exmple, you cn t dd 2 dimensionl vector to 3 dimensionl vector. Algericlly, you multiply vector y numer y multiplying ech component y the numer: k (,) = (k,k). Vectors tht re multiples re sid to e prllel. Exmple. Compute: () (1,5)+(7,19). () (2, 3,8) (16,11,0). (c) 6 (5,3). (d) 2 (2, 1,2)+4 (1, 1,3). () () (c) (d) (1,5)+(7,19) = (1+7,5+19) = (8,24). (2, 3,8) (16,11,0) = (2 16, 3 11,8 0) = ( 14, 14,8). 6 (5,3) = (6 5,6 3) = (30,18). 2 (2, 1,2)+4 (1, 1,3) = (4, 2,4)+(4, 4,12) = (8, 6,16). Here re some properties of vector rithmetic. There is nothing surprising here. Proposition. Let u, v, nd w e vectors (in the sme spce) nd let k e rel numer. () (Associtivity) ( u+ v)+ w = u+( v + w). () (Commuttivity) u+ v = v + u. (c) (Zero vector) The vector 0 with ll0 components stisfies 0+ u = u nd u+ 0 = u. (d) (Additive inverse) The dditive inverse u of u is the vector whose components re the negtives of the components of u. It stisfies u+( u) = 0. (e) (Distriutivity) k ( u+ v) = k u+k v. 4
5 Note: To sy tht the vectors re in the sme spce mens tht, for exmple, u, v, nd w re ll vectors in R 3. But ll of the results re true if u, v, nd w re vectors in R 100 (100dimensionl Eucliden spce). Proof. The ide in ll these cses is to write the vectors in component form nd do the computtion. For exmple, here is proof of (c) in the cse tht u = (u 1,u 2,u 3 ) R u = (0,0,0)+(u 1,u 2,u 3 ) = (u 1,u 2,u 3 ) = u. u+ 0 = (u 1,u 2,u 3 )+(0,0,0) = (u 1,u 2,u 3 ) = u. Here is proof of (e). I ll consider the specil cse where u nd v re vectors in R 3. Thus, Then u = (u 1,u 2,u 3 ) nd v = (v 1,v 2,v 3 ). k ( u+ v) = k [(u 1,u 2,u 3 )+(v 1,v 2,v 3 )] = k (u 1 +v 1,u 2 +v 2,u 3 +v 3 ) = (k (u 1 +v 1 ),k (u 2 +v 2 ),k (u 3 +v 3 )) = (k u 1 +k v 1,k u 2 +k v 2,k u 3 +k v 3 ) = (k u 1,k u 2,k u 3 )+(k v 1,k v 2,k v 3 ) = k (u 1,u 2,u 3 )+k (v 1,v 2,v 3 ) = k u+k v The other prts re proved in similr fshion. There is n lternte nottion for vectors tht is often used in physics nd engineering. î, ĵ, nd ˆk re the unit vectors in the x, y, nd z directions: î = (1,0,0), ĵ = (0,1,0), ˆk = (0,0,1). Note tht For exmple, (,,c) = (1,0,0)+ (0,1,0)+c (0,0,1) = î+ĵ+cˆk. ( 3,6,10) = 3î+6ĵ+10ˆk. In 2 dimensions, (,) = î+ĵ. There is no îĵˆk nottion for vectors with more thn 3 components. You operte with vectors using the îĵˆk nottion in the ovious wys. For exmple, 2(3î 4ˆk)+(5î+7ĵ 11ˆk) = 11î+7ĵ 19ˆk. Geometriclly, multiplying vector y numer multiplies the length of the rrow y the numer. In ddition, if the numer is negtive, the rrow s direction is reversed: 23 5
6 You dd geometric vectors s shown elow. Move one of the vectors sy keeping its length nd direction unchnged so tht it strts t the end of the other vector. Since the copy hs the sme length nd direction s the originl, it s equl to. + Next, drw the vector which strts t the strting point of nd ends t the tip of. This vector is the sum +. The picture elow illustrtes why the geometric ddition rule follows from the lgeric ddition rule. It is oviously specil cse with two 2dimensionl vectors with positive components, ut I think it mkes the result plusile. To dd severl vectors, move the vectors (keeping their lengths nd directions unchnged) so tht they re hedtotil. In the second picture elow, I moved nd c. c c + + c c Finlly, drw vector from the strt of the first vector to the end of the lst vector. Tht vector is the sum in this cse, + + c. The picture elow shows how to sutrct one vector from nother in this cse, is the vector which goes from the tip of to the tip of.  6
7 There re couple of wys to see this. First, if you interpret this s n ddition picture using the hedtotil rule, it sys +( ) =. Alterntively, construct y flipping round, then dd to. This gives +( ) =. As the picture shows, it is the sme s the vector from the hed of to the hed of, ecuse the two vectors re opposite sides of prllelogrm. Exmple. Vectors nd re shown in the picture elow. Drw pictures of the vectors 2, 3 +2, nd 2. c 2017 y Bruce Ikeng 7
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