Find the cartesian equation fo the following planes.vecr.(2hati+3hatj-4hatk)=1
Answer»
ANSWER :Replacing `vecr` by `xhati+yhatj+zhatk`, we have `(xhati+yhatj+zhatz).(2hati+3hatj-4hatk)=1` i.e., `2x+3y-4z =1` which is the cartesian EQUATION of the PLANE.