12
$\begingroup$

I would like to expand a dot product which includes vectors $ \vec{v_1}, \vec{v_2}, \dots $ and constants $ c_1, c_2, \dots $

So that: $$ c_1 \vec{v_1} \cdot \left(c_2 \vec{v_2}+ c_3 \vec{v_3}+\dots \right) = c_1 c_2 \left(\vec{v_1} \cdot \vec{v_2} \right)+ c_1 c_3 \left(\vec{v_1} \cdot \vec{v_3}\right) +\dots $$

I have:

Unprotect[Dot]; 
Dot[a_, Element[d_, Reals] b_, c_] := d Dot[a, b, c] 
Protect[Dot];

Which works for:

$\qquad \left( 3 \vec{v_1}\right)\cdot\left( 4 \vec{v_2}\right) = 12 \vec{v_1}\cdot\vec{v_2} $

But this does not seem to work when I use variables:

$\qquad \mathsf{Element}[c_1 \mid c_2,\, \mathsf{Reals}]$

$\qquad \left( c_1 \vec{v_1}\right)\cdot\left( c_2 \vec{v_2}\right) = \left( c_1 \vec{v_1}\right)\cdot\left( c_2 \vec{v_2}\right) $

How can this be achieved? Ideally, I would like to do this in as general a way as possible.

$\endgroup$
2
  • $\begingroup$ Does Distribute do what you want? $\endgroup$ Oct 30, 2015 at 18:24
  • $\begingroup$ @Patrick Distribute[(c1 v1) . (c2 v2 + c3 v3)] will give (c1 v1).(c2 v2) + (c1 v1).(c3 v3), i.e. not quite there yet. Or where you suggesting something more? $\endgroup$
    – MarcoB
    Oct 30, 2015 at 19:38

3 Answers 3

9
$\begingroup$

Some time ago I developed a function for these sort of symbolic vector computations, called dotExand. For example, it will expand (2 x + 3 y, x-y) to 2(x,x)+(x,y)-3(y,y). Here I use the, at least in my country more common, notation (x,y) for the inner product of the vectors x and y. The advantage is that complicated expressions become more readable in this notation than in the infix notation with the dot. To achieve this output, I define

MakeBoxes[Dot[a_, b_], StandardForm] := RowBox[{"(", ToBoxes[a], ",", ToBoxes[b], ")"}];

The function dotExpand will work both in real and complex vector spaces. To this end, there is a second argument dom, which may take the values Reals or Complexes. The default value is Reals.

dotExpand[exp_, dom : (Reals | Complexes) : Reals] :=
  Module[{res},
    res = ExpandAll[exp] /. Dot[x_, y_] :> Distribute[Dot[x, y]];
    res = res //. {
      Dot[ (a_ /; Simplify[a \[Element] dom])   b_, c_] :> a Dot[b, c],
      Dot[ a_,  (b_ /; Simplify[b \[Element] dom]) c_] :> Simplify[Conjugate[b]]  Dot[a, c],
      If[dom === Reals, Dot[x_, y_] :> Sort[Dot[x, y]], Unevaluated[Sequence[]]]};
    Collect[res, Dot[_, _]]  ]

dotExpand[Dot[ 2 x + 3 y, x - y ]]
(* 2 (x, x) + (x, y) - 3 (y, y) *)

dotExpand[Dot[ 2 x + I y, x - I y ], Complexes]
(* 2 (x, x) + 2 I (x, y) + I (y, x) - (y, y) *)

Your question:

Assuming[ {c1, c2, c3} \[Element] Reals, dotExpand[ Dot[ c1 v1, c2 v2 + c3 v3]]]

(* c1 c2 (v1, v2) + c1 c3 (v1, v3) *)

The complex case:

Assuming[ {c1, c2, c3} \[Element] Complexes, dotExpand[ Dot[ c1 v1, c2 v2 + c3 v3], Complexes]]

(* c1 Conjugate[c2] (v1, v2) + c1 Conjugate[c3] (v1, v3) *)

Finally, consider in a real vector space a triangle with vertices a, b and c such that all three have the same lenght (i.e. the origin is in the centre of the circumscribed circle). Then the orthocentre of this triangle is given by h=a+b+c. The proof of this formula is a matter of substitution:

With[{h = a + b + c}, dotExpand[{(a - h).(c - b) == 0, (b - h).(a - c) == 0}] /. Dot[z_, z_] -> R^2]
(* {True, True} *)
$\endgroup$
4
$\begingroup$

Here is the best I've got so far. If you can be persuaded to write your vectors in a more distinguishable way, e.g. $v_1$ as v[1], then the following might work for you. I'm not so sure how robust it will be, but play around with it and let me know:

Clear[specialDot]
specialDot[expr_] := ReplaceAll[
  Distribute[expr, Plus, Dot, Plus, Times],
  v[a_] v[b_] -> v[a].v[b]
]

Here is the example you had:

specialDot[(c1 v[1]).(c2 v[2] + c3 v[3])]
(* Out: c1 c2 v[1].v[2] + c1 c3 v[1].v[3]*)

and here is another:

specialDot[(a v[1] + b v[2] - c v[3]).(d v[4] - e v[5])]
(* Out: a d v[1].v[4] - a e v[1].v[5] + b d v[2].v[4] - b e v[2].v[5] - 
        c d v[3].v[4] + c e v[3].v[5] *)
$\endgroup$
1
$\begingroup$

Let Subscript[v, 1][u], Subscript[v, 2][u], Subscript[v, 3][u] be the moving frame along the curve .

Define the dot product, cross product relations between the triad \ vectors

Clear[sc];
Clear[sd];
r1 = ((a_  Subscript[v, s_][u])\[Cross](b_  Subscript[v, t_][u]) :> 
    a b Subscript[v, s][u]\[Cross]Subscript[v, t][u]);
r2 = ((a_  Subscript[v, s_][u])\[Cross]( Subscript[v, t_][u]) :> 
    a Subscript[v, s][u]\[Cross] Subscript[v, t][u]);
r3 = (( Subscript[v, s_][u])\[Cross](b_  Subscript[v, t_][u]) :> 
    b Subscript[v, s][u]\[Cross]Subscript[v, t][u]);
r4 = ((a_  Subscript[v, s_][u])\[Cross](b_  Subscript[v, s_][u]) :> 0);
r5 = ((a_  Subscript[v, s_][u])\[Cross]( Subscript[v, s_][u]) :> 0); 
r6 = (( Subscript[v, s_][u])\[Cross](b_  Subscript[v, s_][u]) :> 0);
r13 = ((a_[u]  Subscript[v, s_][u])\[Cross](b_[u]  Subscript[v, t_][
        u]) :> a[u] b[u] Subscript[v, s][u]\[Cross]Subscript[v, t][u]);
r14 = ((a_ [u] Subscript[v, s_][u])\[Cross]( Subscript[v, t_][u]) :> 
    a[u] Subscript[v, s][u]\[Cross] Subscript[v, t][u]);
r15 = (( Subscript[v, s_][u])\[Cross](b_ [u] Subscript[v, t_][u]) :> 
    b[u] Subscript[v, s][u]\[Cross]Subscript[v, t][u]);
r16 = ((a_[u]  Subscript[v, s_][u])\[Cross](b_[u] Subscript[v, s_][
        u]) :> 0);
r17 = ((a_ [u] Subscript[v, s_][u])\[Cross]( Subscript[v, s_][u]) :> 
    0); 
r18 = (( Subscript[v, s_][u])\[Cross](b_ [u] Subscript[v, s_][u]) :> 
    0);
r19 = (Subscript[v, 1][u]\[Cross]Subscript[v, 2][u] :> 
    Subscript[v, 3][u]);
r20 = (Subscript[v, 2][u]\[Cross]Subscript[v, 1][u] :> -Subscript[v, 
      3][u]);
r21 = (Subscript[v, 2][u]\[Cross]Subscript[v, 3][u] :> 
    Subscript[v, 1][u]);
r22 = (Subscript[v, 3][u]\[Cross]Subscript[v, 2][u] :> -Subscript[v, 
      1][u]);
r23 = (Subscript[v, 3][u]\[Cross]Subscript[v, 1][u] :> 
    Subscript[v, 2][u]);
r24 = (Subscript[v, 1][u]\[Cross]Subscript[v, 3][u] :> -Subscript[v, 
      2][u]);
r25 = (Subscript[v, 1][u]\[Cross]Subscript[v, 1][u] :> 0);
r26 = (Subscript[v, 2][u]\[Cross]Subscript[v, 2][u] :> 0);
r27 = (Subscript[v, 3][u]\[Cross]Subscript[v, 3][u] :> 0);
f1 = ((a_  Subscript[v, s_][u]).(b_  Subscript[v, t_][u]) :> 
    a b Subscript[v, s][u].Subscript[v, t][u]);
f2 = ((a_  Subscript[v, s_][u]).( Subscript[v, t_][u]) :> 
    a Subscript[v, s][u]. Subscript[v, t][u]);
f3 = (( Subscript[v, s_][u]).(b_  Subscript[v, t_][u]) :> 
    b Subscript[v, s][u].Subscript[v, t][u]);
f4 = ((a_  Subscript[v, s_][u]).(b_  Subscript[v, s_][u]) :> a b);
f5 = ((a_  Subscript[v, s_][u]).( Subscript[v, s_][u]) :> a); 
f6 = (( Subscript[v, s_][u]).(b_  Subscript[v, s_][u]) :> 0);
f13 = ((a_[u]  Subscript[v, s_][u]).(b_[u]  Subscript[v, t_][u]) :> 
    a[u] b[u] Subscript[v, s][u].Subscript[v, t][u]);
f14 = ((a_ [u] Subscript[v, s_][u]).( Subscript[v, t_][u]) :> 
    a[u] Subscript[v, s][u]. Subscript[v, t][u]);
f15 = (( Subscript[v, s_][u]).(b_ [u] Subscript[v, t_][u]) :> 
    b[u] Subscript[v, s][u].Subscript[v, t][u]);
f16 = ((a_[u]  Subscript[v, s_][u]).(b_[u] Subscript[v, s_][u]) :> 
    a b);
f17 = ((a_ [u] Subscript[v, s_][u]).( Subscript[v, s_][u]) :> a); 
f18 = (( Subscript[v, s_][u]).(b_ [u] Subscript[v, s_][u]) :> b);
f19 = (Subscript[v, 1][u].Subscript[v, 2][u] :> 0);
f20 = (Subscript[v, 2][u].Subscript[v, 1][u] :> 0);
f21 = (Subscript[v, 2][u].Subscript[v, 3][u] :> 0);
f22 = (Subscript[v, 3][u].Subscript[v, 2][u] :> 0);
f23 = (Subscript[v, 3][u].Subscript[v, 1][u] :> 0);
f24 = (Subscript[v, 1][u].Subscript[v, 3][u] :> 0);
f25 = (Subscript[v, 1][u].Subscript[v, 1][u] :> 1);
f26 = (Subscript[v, 2][u].Subscript[v, 2][u] :> 1);
f27 = (Subscript[v, 3][u].Subscript[v, 3][u] :> 1);
sc[expr_] := {Distribute[expr] /. {r1, r2, r3, r4, r5, r6, r13, r14, 
     r15, r16, r17, r18}} /. {r19, r20, r21, r22, r23, r24, r25, r26, 
   r27}
sd[expr_] := {Distribute[expr] /. {f1, f2, f3, f4, f5, f6, f13, f14, 
     f15, f16, f17, f18}} /. {f19, f20, f21, f22, f23, f24, f25, f26, 
   f27}
{Derivative[1][Subscript[v, 1]][u_] := κ[u] Subscript[v, 2][u],
  Derivative[1][Subscript[v, 2]][
    u_] := τ[u] Subscript[v, 3][u] - κ[u] Subscript[v, 1][
      u],
  Derivative[1][Subscript[v, 3]][u_] := -τ[u] Subscript[v, 2][u]};
expr = (κ[u] τ[u] Subscript[v, 3][u] + κ[u] τ[
       u] Subscript[v, 2][u] + κ[u] τ[u] Subscript[v, 1][
       u])\[Cross](g[u] Subscript[v, 1][u] + y[u] Subscript[v, 2][u]);
expr1 = (κ[u] τ[u] Subscript[v, 3][u] + κ[
       u] τ[u] Subscript[v, 2][u] + κ[u] τ[
       u] Subscript[v, 1][u]).(g[u] Subscript[v, 1][u] + 
     y[u] Subscript[v, 2][u]);
sc[expr]
sd[expr1]
{-y[u] κ[u] τ[u] Subscript[v, 1][u] + 
  g[u] κ[u] τ[u] Subscript[v, 2][u] - 
  g[u] κ[u] τ[u] Subscript[v, 3][u] + 
  y[u] κ[u] τ[u] Subscript[v, 3][u]}

{g[u] κ[u] τ[u] + y[u] κ[u] τ[u]}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.