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

1 Answer
Jun 1, 2016

Answer:

#{-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])}#