What is the unit vector that is orthogonal to the plane containing # (29i-35j-17k) # and # (32i-38j-12k) #?

1 Answer
Jan 18, 2017

Answer:

The answer is #=1/299.7〈-226,-196,18〉#

Explanation:

The vector perpendiculatr 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=〈32,-38,-12〉#

Therefore,

#| (veci,vecj,veck), (29,-35,-17), (32,-38,-12) | #

#=veci| (-35,-17), (-38,-12) | -vecj| (29,-17), (32,-12) | +veck| (29,-35), (32,-38) | #

#=veci(35*12-17*38)-vecj(-29*12+17*32)+veck(-29*38+35*32)#

#=〈-226,-196,18〉=vecc#

Verification by doing 2 dot products

#〈-226,-196,18〉.〈29,-35,-17〉=-226*29+196*35-17*18=0#

#〈-226,-196,18〉.〈32,-38,-12〉=-226*32+196*38-12*18=0#

So,

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

The unit vector is

#=1/sqrt(226^2+196^2+18^2)〈-226,-196,18〉#

#=1/299.7〈-226,-196,18〉#