Find the intercepts made on the coordinate axes by the plane 2x + y – 2z = 3 and also find the direction cosines of the normal to the plane.
The given equation of the plane is 2x + y – 2z = 3
Dividing by 3 on both the sides, we get
We know that, if a, b, c are the intercepts by the plane on the coordinate axes, new equation of the plane is
Comparing the equation (i) and (ii), we get
Again the given equation of the plane is
2x+y-2z=3
Writing this in the vector form, we get
So vector normal to the plane is given by
Direction vector of
Direction vector of
So,
Intercepts by the plane on the coordinate axes are
Direction cosines of normal to the plane are