# Plane | Part-2 | Ex-5A

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 