Question

Form the pde representing all planes that are at a constant perpendicular distance d from the origin

Answer 1

Answer:

Let the required equation of the plane be $z$=$lx+my+n\\lx+my-z+n=0$.....(1)Now the plane $(1)$ is at constant distance $a$ from the origintherefore a=$\frac{|n|}{\sqrt{l^2+m^2+1}} \implies a$=$\frac{\pm n}{\sqrt{l^2+m^2+1}} \mbox{Here }p$=$\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \implies n$=$\frac{\pm n}{\sqrt{l^2+m^2+1}}$therefore (1) becomes$lx+my-z\pm a\sqrt{l^2+m^2+1}$=0.....(2)Differentiating $(2)$ with respect to $x$and $y$, we get$l-\frac{dz}{dx}$=$0\mbox{ and }m-\frac{dz}{dy}$=$0 or p$=$l\mbox{ and }q=m \therefore(2)\mbox{ reduces to } px+qy-z\pm a\sqrt{p^2+q^2+1}$=$0 \implies z=px+qy\pm \sqrt{p^2+q^2+1}\mbox{ is the required differential equation}$

Answer 2

  The pde representing all planes that are at a constant perpendicular distance d from the origin is px+qy+a$\sqrt{1+p^2+q^2} = 3$

Given:

Partial differential equation representing all planes that are at a constant perpendicular

To Find:

To Form a partial differential equation

Solution:

lx + my +nz = a

$l^2 + m^2 +n^2 =1$

$n^2 = 1- l^2 -m^2$

lx + my + $\sqrt{1-l^2-m^2} z = a$............(1)

Differentiate (1) partially with respect to x (y const)

= $l= - \sqrt{1 - l^2 - m^2} p$

$p = \frac{-l}{\sqrt{1-l^2-m^2} }$

Differentiating (1) partially with respect to y ,

We have,

q = $\frac{-m}{\sqrt{1-l^2-m^2} }$

$p^2+q^2 =$ $\frac{-l}{\sqrt{1-l^2-m^2} }$ + $\frac{-m}{\sqrt{1-l^2-m^2} }$

After resolving,

$l = \frac{-p}{\sqrt{1+p^2+q^2} }$ and  m = $\frac{-q}{\sqrt{1+p^2+q^2} }$

Sub in (1),

We get ,

Hence , the value of pde is px+qy+a$\sqrt{1+p^2+q^2} = 3$

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