Examples on Equations of Three Dimensional Planes Set 2

Three dimensional planes

Example – 12

A plane is at a distance p from the origin and the direction cosines of the (outward) normal to it are l, m, n. Find its equation.

Solution: The unit vector  \(\hat n\)  normal to the plane is

\[\hat n = l\hat i + m\hat j + n\hat k\]

For any point  \(\vec r\left( {x\hat i + y\hat j + z\hat k} \right)\)  in the plane, we have

\[\begin{align}&\qquad\;\;\;\vec r \cdot \hat n = p \\\\   &\Rightarrow \quad  lx + my + nz = p  \\ \end{align} \]

This is the required equation; it is called the normal form of the plane’s equation. As an exercise, convert the general equation of the plane

\[ax + by + cz + d = 0\]

into normal form.

Example – 13

Find the angle of intersection of the two planes

\[\begin{align}  {a_1}x + {b_1}y + {c_1}z + {d_1} = 0  \\\\  {a_2}x + {b_2}y + {c_2}z + {d_2} = 0  \\ \end{align} \]

Solution: From the equations of the planes, it is evident that the following vectors are to these planes arespectively:

\[\begin{align}  {{\vec n}_1} = {a_1}\hat i + {b_1}\hat j + {c_1}\hat k  \\\\  {{\vec n}_2} = {a_2}\hat i + {b_2}\hat j + {c_2}\hat k  \\ \end{align} \]

Since the acute angle  \(\theta \) between the two planes will be the acute angle between their normals, we have

\[\begin{align}&\cos \theta  = \frac{{\left| {{{\vec n}_1} \cdot {{\vec n}_2}} \right|}}{{\left| {{{\vec n}_1}} \right|\left| {{{\vec n}_2}} \right|}} \\\\  \,\,\,\,\,\,\,\,\,\,\, &\qquad= \frac{{\left| {{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \right|}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }} \\ \end{align} \]

Incidentally, we can now derive the conditions for these planes to be parallel or perpendicular.

Planes are parallel if \({\vec n_1} = \lambda {\vec n_2}\)        \(\begin{align} \Rightarrow \,\,\,\,\,\,\,\,\,\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align}\)

Planes are perpendicular if  \({\vec n_1} \times {\vec n_2} = 0\)         \( \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\)

It should be obvious that for two parallel planes, their equations can be written so that they differ only in the constant term. Thus, any plane parallel to  \(ax + by + cz + d = 0\)  can be written as  \(ax + by + cz + d' = 0\) where   \(d' \in \mathbb{R}\,\,\left( {{\text{and }}d' \ne d} \right).\)

Example – 14

(a) Find the distance of the point  \(P\left( {{x_1},{y_1},{z_1}} \right)\)  from the plane  \(ax + by + cz + d = 0.\)

(b) Find the distance between the two parallel planes

\[\begin{align}  ax + by + cz + {d_1} = 0 \\\\  ax + by + cz + {d_2} = 0  \\\end{align} \]

Solution:(a) The distance l of P from the given plane will obviously be measured along the normal to the plane passing through P:

We write the equation of the plane as

\[\vec r \cdot \vec n =  - d\]

where  \(\vec r = x\hat i + y\hat j + z\hat k\)  is any point on the plane and  \(\vec n = a\hat i + b\hat j + c\hat k\)  is the normal to the plane. Let O be the origin. Since Q lies on the plane, its position vector \(\overrightarrow {OQ} \)  must satisfy the equation of the plane. But  \(\overrightarrow {OQ}  = \overrightarrow {OP}  + \overrightarrow {PQ} .\)    Thus,

\[\left( {\overrightarrow {OP}  + \overrightarrow {PQ} } \right) \cdot \vec n =  - d\]

Note that  \(\overrightarrow {PQ}  = \frac{{\lambda \vec n}}{{\left| {\vec n} \right|}}\)   where   \(\lambda  =  \pm l\)  (which sign to take depends on which direction  \(\vec n\)  points in).

Thus,

\[\begin{align}&\qquad\;\;\;\;\left( {\overrightarrow {OP}  + \frac{{\lambda \vec n}}{{\left| {\vec n} \right|}}} \right) \cdot \vec n =  - d  \\\\   &\Rightarrow  \quad \overrightarrow {OP}  \cdot \vec n + \frac{{\lambda \vec n \cdot \vec n}}{{\left| {\vec n} \right|}} =  - d  \\\\   &\Rightarrow \quad\left( {{x_1}\hat i + {y_1}\hat j + {z_1}\hat k} \right) \cdot \left( {a\hat i + b\hat j + c\hat k} \right) + \lambda \left| {\vec n} \right| =  - d  \\\\   &\Rightarrow  \quad a{x_1} + b{y_1} + c{z_1} + d =  - \lambda \left| {\vec n} \right|  \\   &\Rightarrow  \quad \left| \lambda  \right| = l = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\left| {\vec n} \right|}} \\ \\   &\Rightarrow  \quad l = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}  \\ \end{align} \]

(b) Assume any point  \(P\left( {{x_1},{y_1},{z_1}} \right)\) on the first plane. We have

\[\begin{align}&\qquad\;\;\; a{x_1} + b{y_1} + c{z_1} + {d_1} = 0  \\\\ &\Rightarrow  \quad {d_1} =  - \left( {a{x_1} + b{y_1} + c{z_1}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right) \\ \end{align} \]

The distance of P from the second plane, say l, can be evaluated as described in part (a) above :

\[l = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\]

Using (1) in (2), we have

\[l = \frac{{\left| {{d_2} - {d_1}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

This is the required distance between the two planes.