Find unit vector perpendicular to i-2j+2k and 3i+j-2k ?

1 Answer
Mar 15, 2018

The unit vector is #=1/sqrt117<2,8,7>#

Explanation:

A vector perpendicular to #2# vectors is calculated with the determinant (cross product of 2 vectors)

#| (veci,vecj,veck), (d,e,f), (g,h,i) | #

where #veca=〈d,e,f〉# and #vecb=〈g,h,i〉# are the 2 vectors

Here, we have #veca=〈1,-2,2〉# and #vecb=〈3,1,-2〉#

Therefore,

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

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

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

#=〈2,8,7〉=vecc#

Verification by doing 2 dot products

#〈1,-2,2〉.〈2,8,7〉=(1)*(2)+(-2)*(8)+(2)*(7)=0#

#〈3,1,-2〉.〈2,8,7〉=(3)*(2)+(1)*(8)+(-2)*(7)=0#

So,

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

The unit vector is

#hatc=vecc/||vecc||#

#||vecc||=||<2,8,7>||=sqrt(2^2+8^2+7^2)=sqrt117#

The unit vector is #hatc=1/sqrt117<2,8,7>#