What is the unit vector that is orthogonal to the plane containing # ( - 4 i - 5 j + 2 k) # and # (4 i + 4 j + 2 k) #?

Question

What is the unit vector that is orthogonal to the plane containing #  ( - 4 i - 5 j + 2 k)  # and #  (4 i + 4 j + 2 k)        #?

William Audy · Accepted Answer

The unit vector is #1/sqrt(596)*〈-18,16,4〉#

A vector that is orthogonal to #2# other vectors is calculated with the cross product. The latter is calculate with the determinant.

#| (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=〈-4,-5,2〉# and #vecb=〈4,4,2〉#

Consequently,

#| (veci,vecj,veck), (-4,-5,2), (4,4,2) | #

#=veci| (-5,2), (4,2) | -vecj| (-4,2), (4,2) | +veck| (-4,-5), (4,4) | #

#=veci((-5)*(2)-(4)*(2))-vecj((-4)*(2)-(4)*(2))+veck((-4)*(4)-(-5)*(4))#

#=〈-18,16,4〉=vecc#

Verification using the two dot method

#〈-18,16,4〉.〈-4,-5,2〉=(-18)*(-4)+(16)*(-5)+(4)*(2)=0#

#〈-18,16,4〉.〈4,4,2〉=(-18)*(4)+(16)*(4)+(4)*(2)=0#

Thus,

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

The vector of units is

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

The magnitude of #vecc# is

#||vecc||=||〈-18,16,4〉||=sqrt((-18)^2+(16)^2+(4)^2)#

#=sqrt(596)#

The unit vector is #1/sqrt(596)*〈-18,16,4〉#

Nathan Ashcroft · Answer

undefined The unit vector orthogonal to the plane containing (-4i - 5j + 2k) and (4i + 4j + 2k) is (1/√3)i + (1/√3)j + (1/√3)k.