Saturday, 9 August 2014

CHAPTER 5 - Introduction to Plane

In the chapter on Vectors, we have already learnt how to write the equations for a plane, in different forms. In this section, we will extend that discussion and learn how to write the equation of a plane in three dimensional coordinates form.
The general vector equation of a plane is of the form
\vec r \cdot \vec n = l  : l is a constant
where \vec r is the variable vector x\hat i + y\hat j + z\hat k representing any point on the plane, while \vec n is a fixed vector, say a\hat i + b\hat j + c\hat k which is perpendicular to the plane. Thus, the equation of the plane can be written as
\left( {x\hat i + y\hat j + z\hat k} \right) \cdot \left( {a\hat i + b\hat j + c\hat k} \right) = l
 \Rightarrow  \,\,\,\, ax + by + cz = l
\Rightarrow\,\,\,\,{ax + by + cz + d = 0}\,\,\,\,\,\,\,d =  - l
This is the most general equation of a plane in coordinate form. Note that this equation of the plane contains only three arbitrary constants, for, it can be written as
\left( {\dfrac{a}{d}} \right)x + \left( {\dfrac{b}{d}} \right)y + \left( {\dfrac{c}{d}} \right)z + 1 = 0
 \Rightarrow  \,\,\,\, {\lambda _1}x + {\lambda _2}y + {\lambda _3}z + 1 = 0
Thus, three independent constraints are sufficient to uniquely determine a plane. For example, three non collinear points are sufficient to uniquely determine the plane passing through them.
     Example: 9     

Write the equation of an arbitrary plane passing through the point A\left( {{x_1},{y_1},{z_1}} \right).
Solution: 9

Let us denote the position vector of A  by \vec A\,;\,\vec A is therefore {x_1}\hat i + {y_1}\hat j + {z_1}\hat k.Now, assume that the normal to the plane is \vec n = a\hat i + b\hat j + c\hat k, where abcare variable :
Thus, for any variable point \vec r = x\hat i + y\hat j + z\hat k on the plane, since \left( {\vec r - \vec A} \right) is perpendicular to \vec n, we have
\left( {\vec r - \vec A} \right) \cdot \vec n = 0
 \Rightarrow  \,\,\,\, \left( {\left( {x - {x_1}} \right)\hat i + \left( {y - {y_1}} \right)\hat j + \left( {z - {z_1}} \right)\hat k} \right) \cdot \left( {a\hat i + b\hat j + c\hat k} \right) = 0
 \Rightarrow  \,\,\,\, a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\ldots(1)
This is the required equation of an arbitrary plane through the point A({x_1},{y_1},{z_1}).
We could have arrived at this equation alternatively as follows: we assume the general equation of a plane which is
ax + by + cz + d = 0\ldots(2)
If this passes through \left( {{x_1},{y_1},{z_1}} \right) we have
a{x_1} + b{y_1} + c{z_1} + d = 0\ldots(3)
By \left( 2 \right) - \left( 3 \right), we arrive at the same equation as in  (1).

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