Mark M. answered 05/03/19
Retired Math prof with teaching and tutoring experience in trig.
Let v = J + 4k and w = i - 2j + 3k
The cross product, v x w, is perpendicular to both v and w
v x w is the "determinant" of the 3x3 matrix with first row i j k, second row 0 1 4, and third row 1 -2 3
v x w = 11i + 4j - k
ll v x w ll = magnitude of v x w = √[112 + 42 + (-1)2] = √138
Two unit vectors perpendicular to both v and w are (1 / llv x wll)(v x w) and (-1 / llv x wll)(v x w).