# 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