tg

tg
tgt

Saturday, 9 August 2014

CHAPTER 7 - Worked Out Examples 2

   Example: 13      

Find the angle of intersection of the two planes
{a_1}x + {b_1}y + {c_1}z + {d_1} = 0
{a_2}x + {b_2}y + {c_2}z + {d_2} = 0
Solution: 13

From the equations of the planes, it is evident that the following vectors are to these planes arespectively:
{\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
Since the acute angle \theta  between the two planes will be the acute angle between their normals, we have
\cos \theta  = \dfrac{{\left| {{{\vec n}_1} \cdot {{\vec n}_2}} \right|}}{{\left| {{{\vec n}_1}} \right|\left| {{{\vec n}_2}} \right|}}
 = \dfrac{{\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} }}
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}  \Rightarrow\,\,\,\, \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}
Plames 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 whered' \in \mathbb{R}\,\,\left( {{\rm{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 beween the two parallel lines
ax + by + cz + {d_1} = 0
ax + by + cz + {d_2} = 0
Solution: 14-(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}  = \dfrac{{\lambda \vec n}}{{\left| {\vec n} \right|}} where \lambda  =  \pm l (which sign to take depends on which direction \vec n points in).
Thus,
\left( {\overrightarrow {OP}  + \dfrac{{\lambda \vec n}}{{\left| {\vec n} \right|}}} \right) \cdot \vec n =  - d
 \Rightarrow  \,\,\,\, \overrightarrow {OP}  \cdot \vec n + \dfrac{{\lambda \vec n \cdot \vec n}}{{\left| {\vec n} \right|}} =  - d
 \Rightarrow  \,\,\,\, \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  \,\,\,\, a{x_1} + b{y_1} + c{z_1} + d =  - \lambda \left| {\vec n} \right|
 \Rightarrow  \,\,\,\, \left| \lambda  \right| = l = \dfrac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\left| {\vec n} \right|}}
 \Rightarrow  \,\,\,\, l = \dfrac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}
Solution: 14-(b)

Assume any point P\left( {{x_1},{y_1},{z_1}} \right) on the first plane. We have
a{x_1} + b{y_1} + c{z_1} + {d_1} = 0
 \Rightarrow  \,\,\,\, {d_1} =  - \left( {a{x_1} + b{y_1} + c{z_1}} \right)\ldots(1)
The distance of P from the second plane, say l, can be evaluated as described in part (a)  above :
l = \dfrac{{\left| {a{x_1} + b{y_1} + c{z_1} + {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\ldots(2)
Using  (1)  in (2) we have
l = \dfrac{{\left| {{d_2} - {d_1}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}
This is the required distance between the two planes

Post a Comment