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

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Answer 1 · 2017-04-06T09:44:44

The unit vector is $=〈2/sqrt150,-5/sqrt150,-11/sqrt150〉$

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 $veca=〈-3,1,-1〉$ and $vecb=〈-4,5,-3〉$

Therefore,

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

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

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

$=〈2,-5,-11〉=vecc$

Verification by doing 2 dot products

$〈2,-5,-11〉.〈-3,1,-1〉=-6-5+11=0$

$〈2,-5,-11〉.〈-4,5,-3〉=-8-25+33=0$

So,

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

The unit vector is

$=vecc/(||vecc||)$

$=1/sqrt(4+25+121)〈2,-5,-11〉$

$=1/sqrt150〈2,-5,-11〉$

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