The vector perpendicular to the plane containing 2 vectors is calculated with the determinant (cross product)
#| (veci,vecj,veck), (d,e,f), (g,h,i) | #
where #〈d,e,f〉# and #〈g,h,i〉# are the 2 vectors
Here, we have #veca=〈2,-3,-1〉# and #vecb=〈1,4,-2〉#
Therefore,
#| (veci,vecj,veck), (2,-3,-1), (1,4,-2) | #
#=veci| (-3,-1), (4,-2) | -vecj| (2,-1), (1,-2) | +veck| (2,-3), (1,4) | #
#=veci(-3*-2+1*4)-vecj(-2*2+1*1)+veck(2*4+3*1)#
#=〈10,3,11〉=vecc#
Verification by doing 2 dot products
#〈10,3,11〉.〈2,-3,-1〉=10*2-3*3-11*1=0#
#〈10,3,11〉.〈1,4,-2〉=10*1+3*4-11*2=0#
So,
#vecc# is perpendicular to #veca# and #vecb#
The unit vector is
#=vecc/(||vecc||)#
#=(〈10,3,11〉)/(||〈10,3,11〉||)#
#=1/sqrt(100+9+121)〈10,3,11〉#
#=1/sqrt230〈10,3,11〉#
