What is the unit vector that is normal to the plane containing #(- 3 i + j -k)# and #(- 4i + 5 j - 3k)?

1 Answer
Apr 6, 2017

Answer:

The unit vector is #=〈2/sqrt150,-5/sqrt150,-11/sqrt150〉#

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=〈-3,1,-1〉# and #vecb=〈-4,5,-3〉#

Therefore,

#| (veci,vecj,veck), (-3,1,-1), (-4,5,-3) | #

#=veci| (1,-1), (5,-3) | -vecj| (-3,-1), (-4,-3) | +veck| (-3,1), (-4,5) | #

#=veci(1*-3+1*5)-vecj(-3*-3-1*4)+veck(-3*5+1*4)#

#=〈2,-5,-11〉=vecc#

Verification by doing 2 dot products

#〈2,-5,-11〉.〈-3,1,-1〉=-6-5+11=0#

#〈2,-5,-11〉.〈-4,5,-3〉=-8-25+33=0#

So,

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

The unit vector is

#=vecc/(||vecc||)#

#=1/sqrt(4+25+121)〈2,-5,-11〉#

#=1/sqrt150〈2,-5,-11〉#