Proving Vector Equality without Dot Product - A Simple Method

[square_imp]

Now on another question, that states four points in space have 4 position vectors a,b,c and d relative to O. The question tells you that the line O to the mid-point of AB, BC and CD is perpendicular to the lines AB, BC and CD respectively. The question is asking for me to prove that therefore a^2 = b^2 = c^2 = d^2. Now I have used that fact that the dot product of the perpendicular lines = 0 which ends up proving that the coefficients of the vectors a and b for example equal each other. But is this proof that the actual vectors equal each other? Is there a better way of doing it without using the dot product? Hope I have explained this well enough.Thanks.

 

[square_imp]

Just to clarify, I substituted a = xi + yj + zk and b = ui + vj + wk and proved that x^2 + y^2 + z^2 = u^2 + v^2 + w^2. Is this proof that therefore a^2 = b^2?

 

[Fermat]

Just to clarify, I substituted a = xi + yj + zk and b = ui + vj + wk and proved that x^2 + y^2 + z^2 = u^2 + v^2 + w^2. Is this proof that therefore a^2 = b^2?

What you have shown is that a and b have the same length.

i.e. |a| = |b| or √{(xi)² + (yj)² + (zk)²} = √{(ui)² + (vj)² + (wk)²}

That was all you were being asked to prove. Not that the vectors were equal to each other.
When the question used a², I think that meant the square of the modulus, or length, of the vector a.