A plane can be completely illustrated by denoting two intersecting lines which can be translated into a fixed point A and two nonparallel direction vectors. The position vector of *any* general point ** P** on the plane passing through point A and having

**and is given by the equation**

*direction vectors*## Vector equation of a plane

*Parametric equation of a plane: λ , μ are called a **parameters ** λ,μ *

If **N **is considered to be normal to a given plane, then all other normals to that plane are considered parallel to **N **which are resultantly scalar multiples of **N**., In particular,we can say that there are two normals of length 1:

*Normal/Scalar product form of vector equation of a plane*

Consider a vector n passing through a point A. Only one plane through A can be is perpendicular to the vector. Now consider R being any point on the plane other than A as shown above. Then we can say that

*⇒*

*Cartesian equation of a plane *

*=*

Therefore, the Cartesian form is

where *n*_{1}, *n*_{2} and *n*_{3} are the components of *n* and where *n *is called the normal vector.

Example: Find the equation of the plane passing through the three points P_{1}(1,-1,4), P_{2}(2,7,-1), and P_{3}(5,0,-1).

Hence, consider one point on the plane:

In vector form:

Any non-zero scalar multiples of is also a normal vector of the plane. Therefore, Multiply by -1.

Example: Find the equation of the plane with normal vector containing the point (-2, 3, 4).

Example:Find the distance of the plane = 8 from the origin, and the unit vector perpendicular to the plane.

Example: Find the Cartesian equation of the plane through the point A (1, 1, 1) perpendicular to the vector

Solution:

Example: Show that the following vector is perpendicular to the plane containing the points A(1, 0, 2), B(2, 3, -1) and C(2, 2, -1 ).

Solution:

In conclusion*, n* is a vector that is perpendicular to 2 vectors in the plane so is perpendicular to the plane.