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

1 Answer
Sep 23, 2017

Answer:

The unit vector is #=1/sqrt194〈7,-12,-1〉#

Explanation:

The cross product of 2 vectors 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=〈3,2,-3〉# and #vecb=〈2,1,2〉#

Therefore,

#| (veci,vecj,veck), (3,2,-3), (2,1,2) | #

#=veci| (2,-3), (1,2) | -vecj| (3,-3), (2,2) | +veck| (3,2), (2,1) | #

#=veci(2*2+3*1)-vecj(3*2+3*2)+veck(3*1-2*2)#

#=〈7,-12,-1〉=vecc#

Verification by doing 2 dot products

#〈7,-12,-1〉.〈3,2,-3〉=7*3-12*2+1*3=0#

#〈7,-12,-1〉.〈2,1,2〉=7*2-12*1-1*2=0#

So,

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

The modulus of #vecc# is

#||vecc||=sqrt(7^2+(-12)^2+(-1)^2)=sqrt(49+144+1)=sqrt194#

Therefore,

The unit vector is

#hatc=1/sqrt194〈7,-12,-1〉#