What is the unit vector that is normal to the plane containing #(3i + 4j - k)# and #(2i+ j - 3k)#?

Taking the cross product of given vectors #(3i+4j-k)# & #(2i+j-3k)# as follows

#(3i+4j-k)\times (2i+j-3k)#

#=-11i+7j-5k#

Above vector will be normal to the given vectors hence the unit normal vector

#\hat n=\frac{-11i+7j-5k}{|-11i+7j-5k|}#

#=\frac{-11i+7j-5k}{\sqrt{(-11)^2+7^2+(-5)^2}}#

#=\frac{-11i+7j-5k}{\sqrt{195}}#

hence the unit vector #\hat n# normal to the plane containing given vectors will be

#=\pm(\frac{-11i+7j-5k}{\sqrt{195}})#

The vector normal to a plane containing #2# vectors is given by the cross product of 2 vectors, This is calculated 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=〈3,4,-1〉# and #vecb=〈2,1,-3〉#

Therefore,

#| (veci,vecj,veck), (3,4,-1), (2,1,-3) | #

#=veci| (4,-1), (1,-3) | -vecj| (3,-1), (2,-3) | +veck| (3,4), (2,1) | #

#=veci((4)*(-3)-(1)*(-1))-vecj((3)*(-3)-(2)*(-1))+veck((3)*(1)-(4)*(2))#

#=〈-11,7,-5〉=vecc#

Verification by doing 2 dot products

#〈-11,7,-5〉.〈3,4,-1〉=(-11)*(3)+(7)*(4)+(-5)*(-1)=0#

#〈-11,7,-5〉.〈2,1,-3〉=(-11)*(2)+(7)*(1)+(-5)*(-3)=0#

So,

#vecc# is perpendicular to #veca# and #vecb#

#||vecc||= ||〈-11,7,-5〉|| = sqrt((-11)^2+(7)^2+(-5)^2)#

#=sqrt(121+49+25)#

#=sqrt(195)#

The unit vector is

#hatc=vecc/(||vecc||)=1/sqrt(195)〈-11,7,-5〉#