A vector that is orthogonal to #2# other vectors is calculated with the cross product. The latter is calculate with the determinant.
#| (veci,vecj,veck), (d,e,f), (g,h,i) | #
where #veca=〈d,e,f〉# and #vecb=〈g,h,i〉# are the 2 vectors
Here, we have #veca=〈-4,-5,2〉# and #vecb=〈4,4,2〉#
Therefore,
#| (veci,vecj,veck), (-4,-5,2), (4,4,2) | #
#=veci| (-5,2), (4,2) | -vecj| (-4,2), (4,2) | +veck| (-4,-5), (4,4) | #
#=veci((-5)*(2)-(4)*(2))-vecj((-4)*(2)-(4)*(2))+veck((-4)*(4)-(-5)*(4))#
#=〈-18,16,4〉=vecc#
Verification by doing 2 dot products
#〈-18,16,4〉.〈-4,-5,2〉=(-18)*(-4)+(16)*(-5)+(4)*(2)=0#
#〈-18,16,4〉.〈4,4,2〉=(-18)*(4)+(16)*(4)+(4)*(2)=0#
So,
#vecc# is perpendicular to #veca# and #vecb#
The unit vector is
#hatc=(vecc)/(||vecc||)#
The magnitude of #vecc# is
#||vecc||=||〈-18,16,4〉||=sqrt((-18)^2+(16)^2+(4)^2)#
#=sqrt(596)#
The unit vector is #1/sqrt(596)*〈-18,16,4〉#
