What is the unit vector that is normal to the plane containing $(- 4 i + 5 j - k)$ $(- 4i + 5 j-k)$ and #(2i+ j - 3k)?

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Answer 1 · 2017-11-10T15:31:29

The unit vector is $= < - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}} >$ $=<-1/sqrt3, -1/sqrt3, -1/sqrt3 >$

The normal vector perpendicular to a plane 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 of the plane

Here, we have $veca=〈-4,5,-1〉$ and $vecb=〈2,1,-3〉$

Therefore,

$| (veci,vecj,veck), (-4,5,-1), (2,1,-3) |$

$=veci| (5,-1), (1,-3) | -vecj| (-4,-1), (2,-3) | +veck| (-4,5), (2,1) |$

$=veci(5*-3+1*1)-vecj(4*3+1*2)+veck(-4*1-2*5)$

$=〈-14,-14,-14〉=vecc$

Verification by doing 2 dot products

$〈-14,-14,-14〉.〈-4,5,-1〉=-14*-4+-14*5+14*1=0$

$〈-14,-14,-14〉.〈2,1,-3〉=-28-14+14*3=0$

So,

$vecc$ is perpendicular to $veca$ and $vecb$

$||vecc||=sqrt(14^2+14^2+14^2)=14sqrt3$

The unit vector is

$hatc=1/(||vecc||)vecc=1/(14sqrt3)〈-14,-14,-14〉$

$= <-1/sqrt3, -1/sqrt3, -1/sqrt3 >$

What is the unit vector that is normal to the plane containing (- 4i + 5 j-k)(−4i+5j−k)(- 4i + 5 j-k) and #(2i+ j - 3k)?