ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 5

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 $a$ , $b$ , $c$ are 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

	$\left( {\vec r - \vec A} \right) \cdot \vec n = 0$
	$\Rightarrow \,\,\,\, \left( {\left( {x - {x_1}} \right)\hat i + \left( {y - {y_1}} \right)\hat j + \left( {z - {z_1}} \right)\hat k} \right) \cdot \left( {a\hat i + b\hat j + c\hat k} \right) = 0$
	$\Rightarrow \,\,\,\, a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0$	$\ldots(1)$