What is the unit vector that is orthogonal to the plane containing # (2i + 3j – 7k) # and # (-2i- 3j + 2k) #?

1 Answer
Jun 4, 2017

Answer:

The unit vector is #=<-3/sqrt13, 2/sqrt13,0>#

Explanation:

The vector perpendicular to 2 vectors is calculated with the determinant (cross product)

#| (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=〈2,3,-7〉# and #vecb=〈-2,-3,2〉#

Therefore,

#| (veci,vecj,veck), (2,3,-7), (-2,-3,2) | #

#=veci| (3,-7), (-3,2) | -vecj| (2,-7), (-2,2) | +veck| (2,3), (-2,-3) | #

#=veci(3*2-7*3)-vecj(2*2-7*2)+veck(-2*3+2*3)#

#=〈-15,10,0〉=vecc#

Verification by doing 2 dot products

#〈-15,10,0〉.〈2,3,-7〉=-15*2+10*3-7*0=0#

#〈-15,10,0〉.〈-2,-3,2〉=-15*-2+10*-3-0*2=0#

So,

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

The modulus of #vecc # is #||vecc||=sqrt(15^5+10^2)=sqrt(325)#

The unit vector is

#hatc=vecc/||vecc||=1/325<-15,10,0>#

#=<-3/sqrt13, 2/sqrt13,0>#