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I have an question using vector:

u = {u1, u2}; v = {v1, v2};

$|u|=4$ and $|v|=2$ and $u.v=3$

Find: $(3u-2v).(-u+4v)$

I tried to use some solver, but nothing worked.

Solve[Norm[u] == 4 && Norm[v] == 2 && u.v == 3, {u, v}]

What procedure to follow?

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  • $\begingroup$ The first question that occurs to me is, what dimension are the vectors? That information probably has to be encoded in order to get an answer. One way would be to give u and v explicit components. $\endgroup$ – Michael E2 Apr 15 '18 at 18:59
  • $\begingroup$ (1) You've got three equations and one unknown in the new example problem. (2) Since for all α, the norm of u is greater than 4, there is no solution. (Similar problem with v.) $\endgroup$ – Michael E2 Apr 15 '18 at 19:06
  • $\begingroup$ Doesn't this problem have more solutions than the answers below would suggest? Solve[Norm[u] == 4 && Norm[v] == 2 && u.v == 3, {u \[Element] FullRegion@2, v \[Element] FullRegion@2}, Reals] gives more results... $\endgroup$ – kirma Apr 16 '18 at 4:45
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For dimension is two.

u = {u1, u2}; v = {v1, v2};

Solve[Norm[u] == 4 && Norm[v] == 2 && u.v == 3, {u1, u2, v1, 
    v2}, Reals]

(*   {{u1 -> -(Sqrt[55]/2), u2 -> -(3/2), v1 -> 0, 
       v2 -> -2}, {u1 -> -(Sqrt[55]/2), u2 -> 3/2, v1 -> 0, 
       v2 -> 2}, {u1 -> Sqrt[55]/2, u2 -> -(3/2), v1 -> 0, 
       v2 -> -2}, {u1 -> Sqrt[55]/2, u2 -> 3/2, v1 -> 0, v2 -> 2}}   *)
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  • 1
    $\begingroup$ Ha-ha! You beat me by 4 seconds. :) (+1) $\endgroup$ – Michael E2 Apr 15 '18 at 19:04
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    $\begingroup$ The well known story of the early bird. (+1) $\endgroup$ – Akku14 Apr 15 '18 at 19:08
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If the dimension of the vector is 2:

Thread[List[u, v] -> #] & /@ 
 Block[{u = Array[a, 2], v = Array[b, 2]},
  {u, v} /. 
   Solve[Norm[u] == 4 && Norm[v] == 2 && u.v == 3, Flatten@{u, v}]
  ]
(*
{{u -> {-4, 0}, v -> {-(3/4), -(Sqrt[55]/4)}},
 {u -> {4, 0}, v -> {3/4, -(Sqrt[55]/4)}},
 {u -> {-4, 0}, v -> {-(3/4), Sqrt[55]/4}},
 {u -> {4, 0}, v -> {3/4, Sqrt[55]/4}}}
*)
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