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

1 Answer
Cesareo R.
Jun 1, 2016

{-4 sqrt[2/61], 7/sqrt[122], -3/(sqrt[122])}{−4√261,7√122,−3√122}

Explanation:

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

Their cross product is calculated by the determinant rule, expanding the subdeterminants headed by vec i, vec j, vec k→i,→j,→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])}

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