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
