π,π is the angle FROM a! TO b


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1 Mth 151: 1.2 The Dot Poduct We hve scled vectos (o, multiplied vectos y el nume clled scl) nd dded vectos (in ectngul component fom). Cn we multiply vectos togethe? The nswe is YES! In fct, thee e two diffeent kinds of vecto poducts. One wy of multiplying vectos poduces nothe vecto. This is clled the Coss Poduct, o the Vecto Poduct, (ed vecto coss vecto ). Anothe vecto poduct is the Dot (o Scl) Poduct, (ed vecto dot vecto ), ecuse the poduct is scl (not vecto). This is wht we e studying now. Becuse vectos cn e expessed using eithe pol o ectngul components, thee e two foms y which to wok out the dot poduct. Execise 1: The pol fom of the dot poduct sys to multiply the poduct of the mgnitudes of ech vecto y the cosine of the ngle etween the two vectos. Use the pol fom of the dot poduct to evlute ) î î ) ĵ ĵ c) î ĵ d) ĵ î Execise 2: Show tht!! = x x + y y fo vectos! = x î + y ĵ nd! = x î + y ĵ. The Dot Poduct! Let two vectos e given: =,α o x, y ectngul foms espectively, whee!! = nd = nd! =,β o x, y θ = β α π,π is the ngle FROM! TO! x = cosα nd y = sinα x = cosβ nd y = sinβ in pol o Pol Fom of the Dot Poduct = cosθ Rectngul Fom of the Dot Poduct = x x + y y Theefoe: = cosθ = x x + y y cosθ = x x + y y Relene Gison of 8
2 Mth 151: 1.2 The Dot Poduct Execise 3: Detemine the dot poduct of the vectos! = î 3 ĵ nd! = 4î + 4 ĵ using () the ectngul nd () the pol foms of the dot poduct. Simplify you nswe. Relene Gison of 8
3 Mth 151: 1.2 The Dot Poduct Execise 4: Use!! = cosθ to exploe specil cses of the dot poduct. ) How cn two nonzeo vectos hve zeo dot poduct? ) Wht is the mximum dot poduct two nonzeo vectos of mgnitudes nd cn hve, nd how? c) Cn the dot poduct e negtive? Explin. Execise 5: Is the dot poduct commuttive? (Does!! =!!?) Explin. Becuse of the even symmety of cosine, we will estict the ngle etween the two vectos to e θ 0,π when using the dot poduct. Execise 6: Detemine the ngle etween two vectos of mgnitudes 10 nd 4 if thei dot poduct is 20. Relene Gison of 8
4 Mth 151: 1.2 The Dot Poduct Execise 7: Evlute!! whee! = 2,5 nd! = 5,2. Wht do you notice nd why? Illustte (i) gphiclly nd (ii) nlyticlly using the Dot Poduct fomuls. A common ppliction of the dot poduct is in clculting the (scl quntity) wok (W, mesued in newtonmetes N m, o joules, J) done y foce (F, mesued in newtons, N) on n oject tht the foce cceletes though displcement (D, mesued in metes, m): W =! F! D = FD cosθ Execise 8: A 10 kg lock slides down fictionless 3 mlong mp tht is inclined y 30 o. How much wok is done y the foce of gvity (o the lock s weight) in cceleting the lock down the mp? Complete the digms elow to illustte. Use g = 10 m/s 2. Rel Wold Digm FeeBody Digm Resolution Digm A lock of mss m shown elow cceletes down fictionless mp ngled fom the hoizontl y θ. m θ Relene Gison of 8
5 Mth 151: 1.2 The Dot Poduct Execise 9: A lock of mss m kg slides down fictionless d mlong mp tht is inclined y n ngle θ. The wok done y the foce of gvity in cceleting the lock down the mp is A) md cosθ B) gd cosθ C) mgd cosθ D) mgd sinθ E) mgd tnθ Execise 10: Answe the following fo the unit vecto ˆ = ) The quntity ˆ is (scl o vecto) with mgnitude of. ) The quntity ˆ is in the sme diection s. c) Detemine two unit vectos tht e in the sme line (one pllel nd the othe ntipllel) s = π, e. Since it is ovious tht the mgnitude of ny unit vecto is 1, the mgnitude is not the inteesting fct out this vecto; the impotnt fct out ˆ is its diection. Hence, ˆ is lso known s the diection vecto fo the diection tht ˆ points in, like pent vecto fo ll (scled) vectos tht e pllel to it. This ings us ck to the sic concepts of simil tingles nd why the slope of line is constnt o 3 2 ˆ o Pojections: To poject is to cst shdow onto the gound. In ech cse, poject the fist vecto onto the second vecto y extending line though the second vecto to ct s the gound (o se of the tingle, o the x xis) then sketch ight tingle using the fist vecto s the hypotenuse. Hint: you my find it helpful to otte the ppe so tht the gound is hoizontl. Poject ed onto lue: Poject lue onto ed: Relene Gison of 8
6 Mth 151: 1.2 The Dot Poduct Poject ed onto lue: Poject lue onto ed: Poject ed onto lue: Poject lue onto ed: Resolving Vectos in Pllel nd Pependicul (Othogonl) Components: Lel the ight tingle whose hypotenuse is nd whose ngle of elevtion is θ. Resolve the ed vecto long the lue se vecto into its pllel nd pependicul component vectos. θ θ Let the ed vecto e, the lue vecto e, the geen vecto e c nd the puple vecto e d. Answe the following: c + d = c is pllel to, unless θ is otuse, in which cse c is ntipllel to. ĉ = â OR ĉ = ˆ (cicle the nswe) d is othogonl to When pojecting onto, foms the hypotenuse of the ight tingle, foms the se nd is pllel to nd foms the height nd is pependicul to. c = poj is clled the pojection vecto of the pojection onto d = oth is clled the pojection vecto of the pojection onto Similly, when we esolve the vecto (long the xes given y nd, i.e., like the x nd y xes e given y diection vectos î nd ĵ ), we get the pllel nd pependicul components of. We do this with fee ody digms (nd foce esolution digms) in physics. Relene Gison of 8
7 Mth 151: 1.2 The Dot Poduct Execise 11: Show tht c =. Include sketch in you poof. Execise 12: Show tht c = 2. Why is thee no solute vlue sign? Definitions 1. Vecto Pojection (o pllel pojection vecto) of onto : poj = c = 2 This is just c, the se vecto of the ight tingle. poj is the pllel pojection vecto of onto. 2. Scl Pojection of onto : comp = This scl vlue my e positive, zeo o negtive, depending on the dot poduct vlue nd if the ngle etween nd is cute, ight o otuse. The solute vlue of this quntity is the mgnitude of c. 3. Othogonl Complements of = 1 î + 2 ĵ is nd, oth of which e pependicul to (nd which e ntipllel to ech othe). The mgnitude of,, nd e the sme nd equl The slope of line though nd is m = Δy Δx The slope of line though nd is m = = Δy fo the line Δx The wod complement (like complementy ngles) mens to. If = 3î 4 ĵ, then the two othogonl complement vectos e: 4. Othogonl Pojection of onto : d = c nd d = oth = poj = 2. Relene Gison of 8
8 Mth 151: 1.2 The Dot Poduct Execise 13: Answe the following, whee c is the pllel pojection vecto of onto. ) cosθ > 0 if θ is ; cosθ = 0 if θ is ; cosθ < 0 if θ is ) The quntity cosθ is the pojection of onto. c) The quntity cosθ is the pojection of onto. d) The quntity is the pojection of onto. e) The quntity is the pojection 2 of onto. f) = g) The quntity is the pojection of onto. h) The quntity is the vecto in the diection of, tht when scled y i) poduces c, the pojection of onto. 2 is the vecto pojection of onto. j) Detemine the scl nd vecto pojections of = 3,1 onto = 2,1. Execise 14: Use the vecto pojections to detemine the minimum distnce fom the point P 4,1 ( ) to the line 2x y = 1. Confim using nlyticl geomety. Vecto Method 1. Choose ny two points on the line, O nd Q. 2. Wite uuu = OQ, nd uuu = OP in stndd vecto fom. 3. On the sketch, poject onto. Sketch nd lel these 2 vectos s well s c nd d. 4. The desied minimum distnce is the mgnitude of d, the othogonl pojection vecto. d = oth = poj = 2 minimum distnce is d = Relene Gison of 8
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