91Ó°ÊÓ

The points, $(1,0,0)$ , and $(0,1,0)$ the line $x=y=z$.

A partial differential equation consists of two or more independent variables, an unknown function (depending on those variables), and partial derivatives of the unknown function with respect to the independent variables.

Consider the points as C and D as follows:

If $A=\left(x_1,y_1,z_1\right),B=\left(x_2,y_2,z_2\right)$.

Then the equation of the line $\stackrel{¯}{AB}$is as follows:

If, $C=\left(x_3,y_3,z_3\right),D=\left(x_4,y_4,z_4\right)$.

Then the direction ratios (d.r"s) of $\stackrel{¯}{CD}$:

$x_4-x_3,y_4-y_3,z_4-z_3$

Equation of AB is given as follows:

$\begin{array}{rcl}\frac{x-1}{0-1}& =& \frac{y-0}{1-0}\\ \frac{x-1}{0-1}& =& \frac{z-0}{0-0}\\ \frac{x-1}{0-1}& ⇒& \frac{x-1}{-1}\\ \frac{x-1}{0-1}& =& \frac{y}{1}\end{array}$

Obtain values as follows:

$\begin{array}{rcl}& =& \frac{z}{0}\\ & =& t\end{array}$

Say where -1,1,0 are called d.r:s of $L_1$.

Let, $L_{2:}\frac{x}{1}=\frac{y}{1}=\frac{z}{1}=s$, say (where 1,1,1 are the d.r"s of $L_2$)

Any point on $L_1=(x,y,z)=(-t+1,t,0)=C$, say

Any point on $L_2=(x,y,z)=(s,s,s)=D$, sayd.r"s of $\stackrel{¯}{CD}$:$s+t-1,s-t,s-0$

Formula: Dot product of d.r.:s of two perpendicular lines is 0.

$\begin{array}{rcl}AB& âŠ\yen & L_{\text{l}}\\ & & (s+t-1,s-t,s)·(-1,1,0)\\ -s-t+1+s-t& =& 0\\ -2t+1& =& 0\\ t& =& \frac{1}{2}\\ \stackrel{¯}{CD}& âŠ\yen & L_2\end{array}$

So, required points:

$C=\left(\frac{1}{2},\frac{1}{2},0\right)D=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$

$\begin{array}{rcl}(s+t-1)·1+(s-t)·1+s·1& =& 0\\ 3s-1& =& 0\\ s& =& \frac{1}{3}\\ C& =& (-t+1,t,0)\end{array}$

Calculate further:

$\begin{array}{rcl}C& =& \left(-\frac{1}{2}+1,\frac{1}{2},0\right)\\ C& =& \left(\frac{1}{2},\frac{1}{2},0\right)\\ D& =& (s,s,s)\\ & =& \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)\end{array}$