Question #8bef5

1 Answer
Apr 21, 2017

Answer:

The unit vector is #=〈10/sqrt230,3/sqrt230,11/sqrt230〉#

Explanation:

The vector perpendicular to the plane containing 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=〈2,-3,-1〉# and #vecb=〈1,4,-2〉#

Therefore,

#| (veci,vecj,veck), (2,-3,-1), (1,4,-2) | #

#=veci| (-3,-1), (4,-2) | -vecj| (2,-1), (1,-2) | +veck| (2,-3), (1,4) | #

#=veci(-3*-2+1*4)-vecj(-2*2+1*1)+veck(2*4+3*1)#

#=〈10,3,11〉=vecc#

Verification by doing 2 dot products

#〈10,3,11〉.〈2,-3,-1〉=10*2-3*3-11*1=0#

#〈10,3,11〉.〈1,4,-2〉=10*1+3*4-11*2=0#

So,

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

The unit vector is

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

#=(〈10,3,11〉)/(||〈10,3,11〉||)#

#=1/sqrt(100+9+121)〈10,3,11〉#

#=1/sqrt230〈10,3,11〉#