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Plane, Objectives, Vector Equation of a Plane, Equation of Plane in Normal from, Conversion of Vector Form into Cartesian Form
If we consider any two points in a plane, the line joining these points will lie entirely in the same plane. This is the characteristic of a plane.
Look at Fig. You know that it is a representation of a rectangular box. This has six faces, eight vertices and twelve edges.
The pairs of opposite and parallel faces are
- and
- and
- and
And the sets of parallel edges are given below:
- and
- and
- and
Objectives
After studying this lesson, you will be able to:
Vector Equation of a Plane
A plane is uniquely determined if any one of the following is known:
- Normal to the plane and its distance from the origin is given.
- One point on the plane is given and normal to the plane is also given.
- It passes through three given non collinear points.
Equation of Plane in Normal From
Let the distance of the plane from origin be and let be a unit vector normal to the plane. Consider as position vector of an arbitrary point on the plane.
Since is the perpendicular distance of the plane from the origin and is a unit vector perpendicular to the plane.
Now
is perpendicular to the plane and lies in the plane, therefore
i.e..
i.e..
i.e.. β¦ (1)
Which is the equation of plane in vector from.
Conversion of Vector Form into Cartesian Form
Let be the co-ordinates of the point and be the direction cosines of .
Then
Substituting these value in equation (1) we get
This is the corresponding Cartesian form of equation of plane in normal form.
Example:
Find the distance of the plane from the origin. Also find the direction cosines of the unit vector perpendicular to the plane.
Solution:
The given equation can be written as
Dividing both sides of given equation by we get
i.e..
D. c. s of unit vector normal to the plane are and distance of plane from origin