What is the unit vector that is orthogonal to the plane containing # (20j +31k) # and # (32i-38j-12k) #?

1 Answer
Nov 13, 2017

Answer:

The unit vector is #==1/1507.8<938,992,-640>#

Explanation:

The vector orthogonal to 2 vectros in 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

Here, we have #veca=〈0,20,31〉# and #vecb=〈32,-38,-12〉#

Therefore,

#| (veci,vecj,veck), (0,20,31), (32,-38,-12) | #

#=veci| (20,31), (-38,-12) | -vecj| (0,31), (32,-12) | +veck| (0,20), (32,-38) | #

#=veci(20*-12+38*31)-vecj(0*-12-31*32)+veck(0*-38-32*20)#

#=〈938,992,-640〉=vecc#

Verification by doing 2 dot products

#〈938,992,-640〉.〈0,20,31〉=938*0+992*20-640*31=0#

#〈938,992,-640〉.〈32,-38,-12〉=938*32-992*38+640*12=0#

So,

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

The unit vector is

#hatc=vecc/||vecc||=(<938,992,-640>)/||<938,992,-640>||#

#=1/1507.8<938,992,-640>#