Given a triangle with the vectors $A,B,C$, if you want to determine if it is isosceles, you need to check if two sides have the same length.
My first thought was to calculate the distance between each vector, which involves calculating the norm of the difference vector - that is:
$$|| A - B|| = \sqrt{(A-B)\cdot(A-B)} \\ ||A - C|| = \sqrt{(A-C)\cdot(A-C)} \\ ||B - C|| = \sqrt{(B-C)\cdot(B-C)}$$
And see if two results are equal.
However, that's quite a bit of legwork right there. I was wondering if it would be sufficient to just compare vector differences:
$$A-B\\ A - C\\ B - C$$
and perhaps
$$B-A\\ C-A\\ C-B$$
too.
Which may or may not be more lightweight. So my question is: is comparing the differences between each vector in the triangle enough to determine if it is isosceles?
yes, it is considering that the vectors A,B,C you are talking of are position vectors, and not the vectors representing the sides of the triangle.