2BHH&9capmv3j~<l"  DV0v1 hvVVp p t 9q(Actually, most of the work is in algebraic manipulation which gives us a nice formula in terms of dot products.).)w Su poomerangketchpad p ^Lk$P p (^cЀD p [<z! DV0v1 hvVVp @ 1XGiven vectors V and W, we seek a vector which is the perpendicular projection of V on W.W.@ 0( @1 6ep DV0v1 hvVVpaBL@f DV0v1 hv VVpbC;p  q DV0v1 hvVVpdC@ p   DV0v1 hvVVp2<2@2D2L2P2T2L2P2T[[/This projection will be a multiple of W, right?e of W, right?t?np 2,B[[np :A[\`np p  DV0v1 hvVVp2<2@2D2L2P2T2L2P2T Gak(It's pointed in the same direction as W.W./ NPJ. PaPNvP`v7C0a0D??$GznR0.j0. T  DV0v1 hvVVp2<2@2D2L2P2T2L2P2Tp@`+TdLook at the right triangle formed by a, b, and c (where the line perpendicular to W goes through b).).).p2D2L@2T2L2P2T 0_s  DV0v1 hvVVp2<2@2D2L2P2T2L2P2T@2one side's length is the same as the projection's.s's,,Pp/T0pr  DV0v1 hvVVp2<2@2D2L2P2T2L2P2T` 1NYes, it has length equal to the hypoteneuse time the cosine of the angle at a,00 0 ZX0 J  DV0v1 hvVVp2<2@2D2L2P2T2L2P2T@+cthe length of the projection is |V| cos a. (This is a perfectly good formula, but we'll proceed to write an even better one in terms of the dot product).).2@0`^>T, -?l?{S$-M>jRA@  DV0v1 hvVVp2<2@2D2L2P2T2L2P2T ;cos a is the dot product of UNIT vectors making an angle a,a,@  M Q#R  DV0v1 hvVVp2<2@2D2L2P2T2L2P2TPP=we can make vectors unit vectors by dividing by their length.h.h.P@p44 @@.#Z"  DV0v1 hvVVp2<2@2D2L2P2T2L2P2Te@X@n cos a = (V/|V|)(W/|W|) l|N-@WW@ Ff>pnWpЮ$@p@S@W!U  DV0v1 hvVVp2<2@2D2L2P2T2L2P2T Bp%Gdand the length of the projection is |V| cos a = |V| (V/|V|)(W/|W|) = (V)(W/|W|)Pˠ B Pˠ B0 0?<UP P DV0v1 hvVVp` 2P2T`P78C|So that's the length, and we know the direction, namely that of W. 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