
1. Reduce the equation of the plane to the intercept form and find its intercepts on the coordinate axes.
Solution.
The given equation can be written as
So, the equation (1) represents the intercept form of the plane and the lengths of the intercepts from axis,
axis and
axis are given by
and
2. Find the normal vector to the plane Also, find a unit vector normal to the plane
Solution.
The normal vector to the plane is given by
So, the unit vector normal to the plane is
3. Find the vector equation of the following planes in scalar product form
Solution.
We know that the equation represents a plane passing through a point having position vector
and parallel to vectors
and
[where
and
are scalars]. Here,
and
The given plane is perpendicular to the vector
So, the vector equation of the plane in scalar product form is
Hence, the required vector equation is given by
Solution.
The given plane is perpendicular to the vector
So, the general form of the plane is given by
Solution.
So, the general form of the plane is given by
4. Find the cartesian form of the equation of the following planes :
Solution.
So, the general form of the plane is given by
So, the cartesian form of the equation of the plane is given by
Solution.
So, the general form of the plane is given by
Hence, the cartesian form of the plane is given by