Let #barA = -3hati + hatj - hat k#
Let #barB = hati + hatk#
Let #barC = # the vector perpendicular to both #barA and barB# such that the following is true:
#barC = barA xx barB#
#barC = |
(hati, hatj, hatk, hati, hatj),
(-3,1,-1,-3,1),
(1,0,1,1,0)
|#
#barC = {1(1) - (-1)(0)}hati + {-1(1) - (-3)(1)}hatj + {-3(0) - (1)(1)}hatk#
#barC = hati + 2hatj -hatk#
Check:
#barC*barA = (1)(-3) + (2)(1) + (-1)(-1) = 0#
#barC*barB = (1)(1) + (2)(0) + (-1)(1) = 0#
This checks but #barC# is not a unit vector.
To make it a unit vector we must divide #barC# by its magnitude.
#|barC| = sqrt(1^2 + 2^2 + (-1)^2)#
#|barC| = sqrt(6)#
Dividing by #sqrt(6)# is the same as multiplying by #sqrt(6)/6#
The unit vector is:
#hatC = sqrt(6)/6hati + sqrt(6)/3hatj - sqrt(6)/6hatk#
