>you will get the same results using a simple matmul, because euclidean distance over normalized vectors is a linear transform of the cosine distance. Squared

>you will get the same results using a simple matmul, because euclidean distance over normalized vectors is a linear transform of the cosine distance. Squared euclidean distance of normalized vectors is an affine transform of their cosine similarity (the cosine of the angle between them). EuclideanDistance(x, y) = sqrt(dot(x - y, x - y)) = sqrt(dot(x, x) - 2dot(x, y) + dot(y, y)) = sqrt(2 - 2dot(x, y))