silemus.

48Gram-Schmidt second vector

Apply Gram–Schmidt to v1=(1,1,0)\displaystyle v_1 = (1, 1, 0), v2=(1,0,1)\displaystyle v_2 = (1, 0, 1). Find the second orthogonal vector u2\displaystyle u_2 before normalising.

Keep u1 = v1 = (1,1,0). Then u2 = v2 minus the part of v2 lying along u1: u2 = (1,0,1) − ½(1,1,0) = (½, −½, 1). Checking, u2·u1 = 0, so they're perpendicular.

How good is this answer?