Question

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

Answer 1

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

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=〈-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〉`