Question

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

Answer 1

`{-4 sqrt[2/61], 7/sqrt[122], -3/(sqrt[122])}`

Explanation:

Given two non aligned vectors `vec u` and `vec v` the cross product given by `vec w = vec u times vec v` is orthogonal to `vec u` and `vec v`

Their cross product is calculated by the determinant rule, expanding the subdeterminants headed by `vec i, vec j, vec k`

`vec w =vec u times vec v =det ((vec i,vec j,vec k),(u_x,u_y,u_z),(v_x,v_y,v_z))`

`vec u times vec v= (u_y v_z-u_z v_y)vec i -(u_xv_z-u_z v_x)vec j + (u_x v_y-u_y v_x)vec k`

so
`vec w =det ((vec i,vec j,vec k),(1,2,2),(2,1,-3)) =-8 vec i+7 vecj-3vec k`

Then the unit vector is `vec w/norm(vec w) = {-4 sqrt[2/61], 7/sqrt[122], -3/(sqrt[122])}`