Vector equation

Question

Is there any way to solve below equation in Mathematica? I want to find vector 'X' as a function of vectors: A,V, and constants: b, e and d. All vectors are 1 by 3. Equation is:

A + b V X.X + X (d V.X + 1 + e X.X) == 0

Which 0 in right-hand side is a 3 component vector. It is not a linear equation. The code I used is as below:

A = {a1, a2, a3};
v = {v1, v2, v3};
x = {x1, x2, x3};
Solve[-A.A + (-2 n - v.v ) (c + b x.x) + x.x (e f x.x + d v.x)^2 == 
 0, x, Real]

Answer 1

This is not an answer, but too long for a comment. I can't manage to solve the question myself, but here are some things:

• In Mathematica, the three-component zero-vector is {0,0,0}. You should replace 0 with this.

• Unless otherwise specified, Mathematica will assume that symbols are complex numbers, not real vectors. What you should do is something like this:

  V = {v1, v2, v3}; A = {a1, a2, a3}; X = {x1, x2, x3};

Now when you type the expression A + b V X.X + X (d V.X + 1 + e X.X) and evaluate it, you will see that Mathematica recognises it as a three-component vector:

 {a1 + b v1 (x1^2 + x2^2 + x3^2) + 
  x1 (1 + d (v1 x1 + v2 x2 + v3 x3) + e (x1^2 + x2^2 + x3^2)), 

 a2 + b v2 (x1^2 + x2^2 + x3^2) + 
  x2 (1 + d (v1 x1 + v2 x2 + v3 x3) + e (x1^2 + x2^2 + x3^2)), 

 a3 + b v3 (x1^2 + x2^2 + x3^2) + 
  x3 (1 + d (v1 x1 + v2 x2 + v3 x3) + e (x1^2 + x2^2 + x3^2))}

Remember: each of these symbols will be assumed by Mathematica to be a scalar, and that's what we want.

• To stop Mathematica giving you complex solutions and to speed up the solving, you should specify that you want to solve over the Reals:

  Solve[ X^2 (d V.X + e f X^2)^2 - (a^2 + V^2 (b X^2 + c) + 2 a.V (b X^2 + c)) == {0,0,0}, X, Reals]

Note that solving for X here happens to work because X evaluates to {x1,x2,x3} which is the list of variables to solve for. (See the documentation for Solve)

Unfortunately, even after this syntax is all corrected, Mathematica seems to have trouble solving this equation. Hopefully someone else can help with that.