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

1 Answer
Feb 19, 2017

Answer:

The unit vector is #=<-2/sqrt30,-1/sqrt30,5/sqrt30>#

Explanation:

We calculate the vector that is perpendicular to the other 2 vectors by doing a cross product,

Let #veca=<-3,1,-1>#

#vecb=<-2,-1,-1>#

#vecc=|(hati,hatj,hatk),(-3,1,-1),(-2,-1,-1)|#

#=hati|(1,-1),(-1,-1)|-hatj|(-3,-1),(-2,-1)|+hatk|(-3,1),(-2,-1)|#

#=hati(-2)-hatj(1)+hatk(5)#

#=<-2,-1,5>#

Verification

#veca.vecc=<-3,1,-1>.<-2,-1,5>=6-1-5=0#

#vecb.vecc=<-2,-1,-1>.<-2,-1,5>=4+1-5=0#

The modulus of #vecc=||vecc||=||<-2,-1,5>||=sqrt(4+1+25)=sqrt30#

The unit vector # = vecc /(||vecc||)#

#=1/sqrt30<-2,-1,5>#