What is the unit vector that is orthogonal to the plane containing # (3i - j - 2k) # and # (3i – 4j + 4k) #?

1 Answer
Jan 8, 2018

Answer:

The unit vector is #=1/sqrt(549)(-12i-18j-9k)#

Explanation:

A vector perpendicular to 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,-1,-2〉# and #vecb=〈3,-4,4〉#

Therefore,

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

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

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

#=〈-12,-18,-9〉=vecc#

Verification by doing 2 dot products

#〈3,-1,-2〉.〈-12,-18,-9〉=-3*12+1*18+2*9=0#

#〈3,-4,4〉.〈-12,-18,-9〉=-3*12+4*18-4*9=0#

So,

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

The unit vector #hatc# in the direction of #vecc# is

#hatc=(vecc)/sqrt((-12)^2+(-18)^2+(-9)^2)=vecc/sqrt(549)#

#=1/sqrt(549)(-12i-18j-9k)#