# Getting False for an equality that I know it is true [closed]

I am getting false for an equality that I know it is true(I proved it by hand and I used sagemath to see that it is true) however when using Mathematica it is giving false although I used TrueQ with Simplify. The equality that I want to check is the following: xi and H are vectors having 3 components

B = Cross[xi, H]
b = B.B,
TrueQ[(H.xi)^2*(xi.xi) == (H.H)*(xi.xi)^2 - b*(xi.xi)//Simplify]


And the answer is False!

• Your code maybe B = Cross[xi, H]; b = B . B; TrueQ[(H . xi)^2*(xi . xi) == (H . H)*(xi . xi)^2 - b*(xi . xi) // Simplify] Jul 18 at 10:10
• The TrueQ command is not an appropriate command to this end. The documentation clear says " TrueQ will return True only if the input is explicitly True" For example, TrueQ[(a + b)^2 == a^2 + 2*a*b + b^2] returns False. Jul 18 at 10:18

An alternative answer using components for the vectors:

xi = {x1, x2, x3};
H = {H1, H2, H3};
B = Cross[xi, H]; b = B.B;
(H.xi)^2*(xi.xi) - ((H.H) (xi.xi)^2 - b (xi.xi)) // FullSimplify


returns

0

hence the identity is true. In principle the vectors could also be complex as the components are unspecified.

$Assumptions = xi ∈ Vectors[3, Reals] && H ∈ Vectors[3, Reals] B = Cross[xi, H]; b = B . B; (H . xi)^2*(xi . xi) == (H . H) (xi . xi)^2 - b (xi . xi) // TensorReduce  True • +1 You can also do this with complex components as well as with symbolic dimensions, $Assumptions = xi \[Element] Vectors[dim] && H \[Element] Vectors[dim]; Jul 18 at 10:55