Question

What is the unit vector that is normal to the plane containing <2i+7j-2k> and <8i-2j+3k>?

Answer 1

The unit vector is `=1/sqrt(4373)*〈17,-22,-60〉`

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=〈2,7,-2〉` and `vecb=〈8,-2,3〉`

Therefore,

`| (veci,vecj,veck), (2,7,-2), (8,-2,3) |`

`=veci| (7,-2), (-2,3) | -vecj| (2,-2), (8,3) | +veck| (2,7), (8,-2) |`

`=veci(7*3-2*2)-vecj(2*3+8*2)+veck(2*-2-7*8)`

`=〈17,-22,-60〉=vecc`

Verification by doing 2 dot products

`〈17,-22,-60〉.〈2,7,-2〉=17*2-22*7-60*-2=0`

`〈17,-22,-60〉.〈8,-2,3〉=17*8-22*-2-60*3=0`

So,

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

The unit vector is

`hatc=vecc/||c||=1/sqrt(4373)*〈17,-22,-60〉`