Question

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

Answer 1

The answer is `=<4/sqrt90,5/sqrt90,-7/sqrt90>`

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=〈1,2,2〉`

Therefore,

`| (veci,vecj,veck), (-3,1,-1), (1,2,2) |`

`=veci| (1,-1), (2,2) | -vecj| (-3,-1), (1,2) | +veck| (-3,1), (1,2) |`

`=veci(1*2+1*2)-vecj(-3*2+1*1)+veck(-3*2-1*1)`

`=〈4,5,-7〉=vecc`

Verification by doing 2 dot products

`〈4,5,-7〉.〈-3,1,-1〉=-12+5+7=0`

`〈4,5,-7〉.〈1,2,2〉=4+10-14=0`

So,

`vecc` is perpendicular to `veca` and `vecb`

The unit vector is

`=1/sqrt(16+25+49)*<4,5,-7>`

`=<4/sqrt90,5/sqrt90,-7/sqrt90>`