What is the unit vector that is orthogonal to the plane containing # (29i-35j-17k) # and # (41j+31k) #?

1 Answer
Apr 2, 2017

Answer:

The unit vector is #=1/1540.3〈-388,-899,1189〉#

Explanation:

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

#| (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=〈29,-35,-17〉# and #vecb=〈0,41,31〉#

Therefore,

#| (veci,vecj,veck), (29,-35,-17), (0,41,31) | #

#=veci| (-35,-17), (41,31) | -vecj| (29,-17), (0,31) | +veck| (29,-35), (0,41) | #

#=veci(-35*31+17*41)-vecj(29*31+17*0)+veck(29*41+35*0)#

#=〈-388,-899,1189〉=vecc#

Verification by doing 2 dot products

#〈-388,-899,1189〉.〈29,-35,-17〉=-388*29+899*35-17*1189=0#

#〈-388,-899,1189〉.〈0,41,31〉=-388*0-899*41+1189*31=0#

So,

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

The unit vector in the direction of #vecc# is

#=vecc/||vecc||#

#||vecc||=sqrt(388^2+899^2+1189^2)=sqrt2372466#

The unit vector is #=1/1540.3〈-388,-899,1189〉#