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

1 Answer
Feb 15, 2017

Answer:

The unit vector is #=<-10/sqrt117,1/sqrt117,4/sqrt117>#

Explanation:

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

Let #veca=<1,-2,3>#

#vecb=<1,2,2>#

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

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

#=hati(-10)-hatj(-1)+hatk(4)#

#=<-10,1,4>#

Verification

#veca.vecc=<1,-2,3>.<-10,1,4>=-10-2+12=0#

#vecb.vecc=<1,2,2>.<-10,1,4>=-10+2+8=0#

The modulus of #vecc=||vecc|=||<-10,1,4>||=sqrt(100+1+16)=sqrt117#

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

#=1/sqrt117<-10,1,4>#