A vector perpendicular to 2 vectors 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
Here, we have #veca=〈3,-1,-2〉# and #vecb=〈3,-4,4〉#
Therefore,
#| (veci,vecj,veck), (3,-1,-2), (3,-4,4) | #
#=veci| (-1,-2), (-4,4) | -vecj| (3,-2), (3,4) | +veck| (3,-1), (3,-4) | #
#=veci(-1*4-(-2)*-4)-vecj(3*4-3*-2)+veck(-4*3-3*-1)#
#=〈-12,-18,-9〉=vecc#
Verification by doing 2 dot products
#〈3,-1,-2〉.〈-12,-18,-9〉=-3*12+1*18+2*9=0#
#〈3,-4,4〉.〈-12,-18,-9〉=-3*12+4*18-4*9=0#
So,
#vecc# is perpendicular to #veca# and #vecb#
The unit vector #hatc# in the direction of #vecc# is
#hatc=(vecc)/sqrt((-12)^2+(-18)^2+(-9)^2)=vecc/sqrt(549)#
#=1/sqrt(549)(-12i-18j-9k)#
