DERIVE for Windows version 5.04 DfW file saved on 20 Jun 2005 u:=[2, 3, 4] v:=[5, -1, 2] #CTextObj {\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Reference Sheet for Quiz #5 -- The MERC Lab Technology Quiz \par \par The assignment operator [colon + equals] allows us to \par automatic substitutions for algebraic expression or functions. \par \par For example, we will let \par u := [2,3,4] and v := [5,-1,2]. \par } CExpnObj8User u:=[2,3,4]8User v:=[5,-1,2]{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 If we Author the expression \par 3u + 4v \par Derive will automatically substitute in the vector values. \par } 8User3*u+4*v 2{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 We click on "Simplify" [equals sign icon]. \par } >8JSimp(#3) [26,5,20]Vz{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 We can find the magnitude of u. \par Use the "Pipes". [Shift the key with the backslash on it.] \par } 8PUserABS(u){\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 "Simplify" it. \par } Simp(#5)SQRT(29){\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Let's create the unit vector associated with v. \par } 8`Userv/ABS(v)"hFSimp(#7)%[SQRT(30)/6,-SQRT(30)/30,SQRT(30)/15]Rv{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 The dot product of u and v is simply \par u*v. Treat it as multiplication. \par } 8PUser u*vSimp(#9) 15}{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 \par } {\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 The orthogonal projection of u onto v is: \par } 8pUser  u*v/(v*v)*vHB Simp(#11)  [5/2,-1/2,1]N;{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 The projection of u ORTHOGONAL to v is \par u - #12. \par We want to subtract the vector on Line #12 from u. \par The pound sign allows us to substitute the contents of \par other lines. \par } 8User u-[5/2,-1/2,1]H Simp(#13) [-1/2,7/2,3]8{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Let's be sure that Line #12 is orthogonal to Line #14... \par #12*#14 \par } 8D0hUser[5/2,-1/2,1]*[-1/2,7/2,3]t Simp(#15)0V{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 So we have decomposed u into components parallel to v \par and orthogonal to v! \par \par Here's the cross product... \par The name of the Derive function is "cross(?,?)". \par The arguments must be 3-D vectors! \par } 8User CROSS(u,v)@( Simp(#17) [10,16,-17]4F{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Is this orthogonal to both u and v? Yes. \par } 8R^User [10,16,-17]*ujv Simp(#19)08User [10,16,-17]*v Simp(#21)0