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

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Answer 1 · 2017-03-25T12:41:20

The answer is #=<4/sqrt90,5/sqrt90,-7/sqrt90>#

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=〈1,2,2〉#

Therefore,

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

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

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

#=〈4,5,-7〉=vecc#

Verification by doing 2 dot products

#〈4,5,-7〉.〈-3,1,-1〉=-12+5+7=0#

#〈4,5,-7〉.〈1,2,2〉=4+10-14=0#

So,

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

The unit vector is

#=1/sqrt(16+25+49)*<4,5,-7>#

#=<4/sqrt90,5/sqrt90,-7/sqrt90>#