ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 7

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 $ax + by + cz + d' = 0$ where $d' \in \mathbb{R}\,\,\left( {{\rm{and }}\,d' \ne d} \right).$

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

Saturday, 9 August 2014

CHAPTER 7 - Worked Out Examples 2

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