Begin by computing the cross product. I use a determinant:
#barA xx barB =
|
(hati, hatj, hatk),
(1, 3, 0),
(2,0, 5)
|#
#barA xx barB = hati|(3, 0),(0,5)| + hatj|(0,1),(5,2)| + hatk|(1,3),(2,0)|#
#barA xx barB = 15hati - 5hatj -6hatk#
Let #barC = 15hati - 5hatj -6hatk#
The unit vector, #hatC = barC/|barC|#
#|barC| = sqrt(15^2 + (-5^2) + (-6)^2)#
#|barC| = sqrt(286)#
#hatC = 15/sqrt(286)hati - 5/sqrt(286)hatj -6/sqrt(286)hatk#
The only other vector that can be orthogonal to #barA and barB# is:
#barB xx barA#
Because #barA xx barB = -(barB xx barA)#, the only other unit vector orthogonal to #barA and barB# is:
#-hatC = -15/sqrt(286)hati + 5/sqrt(286)hatj +6/sqrt(286)hatk#