What is the unit vector that is normal to the plane containing <2i+7j-2k> and <8i-2j+3k>?

1 Answer
May 13, 2017

Answer:

The unit vector is #=1/sqrt(4373)*〈17,-22,-60〉#

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=〈2,7,-2〉# and #vecb=〈8,-2,3〉#

Therefore,

#| (veci,vecj,veck), (2,7,-2), (8,-2,3) | #

#=veci| (7,-2), (-2,3) | -vecj| (2,-2), (8,3) | +veck| (2,7), (8,-2) | #

#=veci(7*3-2*2)-vecj(2*3+8*2)+veck(2*-2-7*8)#

#=〈17,-22,-60〉=vecc#

Verification by doing 2 dot products

#〈17,-22,-60〉.〈2,7,-2〉=17*2-22*7-60*-2=0#

#〈17,-22,-60〉.〈8,-2,3〉=17*8-22*-2-60*3=0#

So,

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

The unit vector is

#hatc=vecc/||c||=1/sqrt(4373)*〈17,-22,-60〉#