What is the unit vector that is orthogonal to the plane containing # (2i + 3j – 7k) # and # (3i – 4j + 4k) #?

1 Answer
Apr 10, 2017

Answer:

The unit vector is #=〈-16/sqrt1386,-29/sqrt1386,-17/sqrt1386〉#

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=〈2,3,-7〉# and #vecb=〈3,-4,4〉#

Therefore,

#| (veci,vecj,veck), (2,3,-7), (3,-4,4) | #

#=veci| (3,-7), (-4,4) | -vecj| (2,-7), (3,4) | +veck| (2,3), (3,-4) | #

#=veci(3*4-7*4)-vecj(2*4+7*3)+veck(-2*4-3*3)#

#=〈-16,-29,-17〉=vecc#

Verification by doing 2 dot products

#〈-16,-29,-17〉.〈2,3,-7〉=-16*2-29*3-7*17=0#

#〈-16,-29,-17〉.〈3,-4,4〉=-16*3+29*4-17*4=0#

So,

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

The unit vector is

#=vecc/||vecc||=1/sqrt(16^2+29^2+17^2)〈-16,-29,-17〉#

#=1/sqrt1386〈-16,-29,-17〉#