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

1 Answer
Jun 11, 2018

Answer:

The unit vector is #=1/sqrt(2870)〈17,-30,-41〉#

Explanation:

First calculate the vector orthogonal to the other #2# vectors. This is given by the cross product.

#| (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=〈-5,4,-5〉#

Therefore,

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

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

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

#=〈17,-30,-41〉=vecc#

Verification by doing 2 dot products

#〈17,-30,-41〉.〈-4,-5,2〉=(17)*(-4)+(-30)*(-5)+(-41)*(2)=0#

#〈17,-30,-41〉.〈-5,4,-5〉=(17)*(-5)+(-30)*(4)+(-41)*(-5)=0#

So,

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

The unit vector is

#hatc=vecc/(||vecc||)=1/sqrt(17^2+(-30)^2+(-41)^2)*〈17,-30,-41〉#

#=1/sqrt(2870)〈17,-30,-41〉#