What is the unit vector that is normal to the plane containing #(- 4i + 5 j-k)# and #(2i+ j - 3k)?

1 Answer
Nov 10, 2017

Answer:

The unit vector is #=<-1/sqrt3, -1/sqrt3, -1/sqrt3 >#

Explanation:

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 >#