The normal vector perpendicular to a plane is calculated with the determinant
#| (veci,vecj,veck), (d,e,f), (g,h,i) | #
where #〈d,e,f〉# and #〈g,h,i〉# are the 2 vectors of the plane
Here, we have #veca=〈-4,5,-1〉# and #vecb=〈2,1,-3〉#
Therefore,
#| (veci,vecj,veck), (-4,5,-1), (2,1,-3) | #
#=veci| (5,-1), (1,-3) | -vecj| (-4,-1), (2,-3) | +veck| (-4,5), (2,1) | #
#=veci(5*-3+1*1)-vecj(4*3+1*2)+veck(-4*1-2*5)#
#=〈-14,-14,-14〉=vecc#
Verification by doing 2 dot products
#〈-14,-14,-14〉.〈-4,5,-1〉=-14*-4+-14*5+14*1=0#
#〈-14,-14,-14〉.〈2,1,-3〉=-28-14+14*3=0#
So,
#vecc# is perpendicular to #veca# and #vecb#
#||vecc||=sqrt(14^2+14^2+14^2)=14sqrt3#
The unit vector is
#hatc=1/(||vecc||)vecc=1/(14sqrt3)〈-14,-14,-14〉#
#= <-1/sqrt3, -1/sqrt3, -1/sqrt3 >#
