91Ó°ÊÓ

Plane ax + by + c z = d, N = (a, b ,c)

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}, \\ D(x,y)={(x-x_0)}^2+{(y-y_0)}^2+{(\frac{d-ax-by}{c}-z_0)}^2....\left(1\right) \\ \mathrm{First}\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{point}, \\ \left(\frac{ad-b{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right) \\ \mathrm{gives}\mathrm{absolute}\mathrm{minimum}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{function}. \\ \mathrm{Differentiate}\left(1\right) \mathrm{partially}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x}, \\ \frac{∂}{∂x}D(x,y)=\frac{∂}{∂x}\{{(x-x_0)}^2+{(y-y_0)}^2+{(\frac{d-ax-by}{c}-z_0)}^2\} \\ {D}_x(x,y)=\frac{∂}{∂x}{(x-x_0)}^2+\frac{∂}{∂x}{(y-y_0)}^2+\frac{∂}{∂x}{(\frac{d-ax-by}{c}-z_0)}^2. \\ =2(x-x_0)+2(\frac{d-ax-by}{c}-z_0)\frac{∂}{∂x}(\frac{d-ax-by}{c}-z_0). \\ =2(x-x_0)+2(\frac{d-ax-by}{c}-z_0)\{\frac{∂}{∂x}\left(\frac{d-ax-by}{c}\right)-\frac{∂}{∂x}{z}_0\} \\ =2(x-x_0)+2(\frac{d-ax-by}{c}-z_0)(\frac{-a}{c}-0)=2(x-x_0)-2\frac{a}{c}(\frac{d-ax-by}{c}-z_0)......\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Differentiate}\left(1\right) \mathrm{partially}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{y}, \\ \frac{∂}{∂y}D(x,y)=\frac{∂}{∂y}\{{(x-x_0)}^2+{(y-y_0)}^2+{(\frac{d-ax-by}{c}-z_0)}^2\} \\ {D}_y(x,y)=\frac{∂}{∂y}{(x-x_0)}^2+\frac{∂}{∂y}{(y-y_0)}^2+\frac{∂}{∂y}{(\frac{d-ax-by}{c}-z_0)}^2. \\ =2(y-y_0)+2(\frac{d-ax-by}{c}-z_0)\frac{∂}{∂y}(\frac{d-ax-by}{c}-z_0). \\ =2(y-y_0)+2(\frac{d-ax-by}{c}-z_0)\{\frac{∂}{∂y}\left(\frac{d-ax-by}{c}\right)-\frac{∂}{∂y}{z}_0\} \\ =2(y-y_0)+2(\frac{d-ax-by}{c}-z_0)(\frac{-b}{c}-0)=2(x-x_0)-2\frac{b}{c}(\frac{d-ax-by}{c}-z_0).....\left(3\right) \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{stationary}\mathrm{point}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {D}_x(x,y)=0and{D}_y(x,y)=0 \\ Thus, \\ {D}_x(x,y)=0 \\ 2(x-x_0)-2\frac{a}{c}(\frac{d-ax-by}{c}-z_0)=0 \\ 2x-2x_0-\frac{2ad}{c^2}+\frac{2a^{2}x}{c^2}+\frac{2aby}{c^2}+\frac{2a{z}_0}{c^{}}=0 \\ \left(2+\frac{2a^2}{c^2}\right)x+\frac{2aby}{c^2}=2x_0+\frac{2ad}{c^2}-\frac{2a{z}_0}{c^{}}......\left(4\right) \\ And, \\ {D}_y(x,y)=0 \\ 2(y-y_0)-2\frac{b}{c}(\frac{d-ax-by}{c}-z_0)=0 \\ 2y-2y_0-\frac{2bd}{c^2}+\frac{2abx}{c^2}+\frac{2b^{2}y}{c^2}+\frac{2b{z}_0}{c^{}}=0 \\ \frac{2abx}{c^2}+\left(2+\frac{2b^2}{c^2}\right)y=2y_0+\frac{2bd}{c^2}-\frac{2b{z}_0}{c^{}}.....\left(5\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Multiply}\left(4\right)\mathrm{by}\frac{2\mathrm{zc}}{{\mathrm{c}}^2}\mathrm{and}\mathrm{multiply}\left(5\right)\mathrm{by}\left(2+\frac{2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)\mathrm{and}\mathrm{then}\mathrm{subtract}, \\ \begin{array}{r}\frac{2ab}{c^2}\left\{\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y\right\}−\left(2+\frac{2a^2}{c^2}\right)\left\{\frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)y\right\}\\ =\frac{2ab}{c^2}\left(2x_0+\frac{2ad}{c^2}−\frac{2a}{c}{z}_0\right)−\left(2+\frac{2a^2}{c^2}\right)\left(2y_0+\frac{2bd}{c^2}−\frac{2b}{c}{z}_0\right)\end{array} \\ \begin{array}{r}\frac{2ab}{c^2}\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}.\frac{2ab}{c^2}y−\left(2+\frac{2a^2}{c^2}\right)\frac{2ab}{c^2}x+\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)y\\ =\frac{4ab}{c^2}{x}_0+\frac{4a^{4}bd}{c^4}−\frac{4a^{2}b}{c^3}{z}_0−4y_0-4\frac{bd}{c^2}\\ +\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0-\frac{4a^{2}bd}{c^4}+\frac{4a^{2}b}{c^3}{z}_0\\ \\ \\ \\ \end{array} \end{gathered}$$

$\begin{array}{r}\left\{\frac{4a^2{b}^2}{c^4}-\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)\right\}y=\frac{4ab}{c^2}{x}_0−4y_0-4\frac{bd}{c^2}+\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0\\ \left(\frac{-4c^2-4a^2-4b^2}{c^2}\right)y=\frac{-4bd+4ab{x}_0+4bc{z}_0-4c^2{y}_0-4a^2{y}_0}{c^2}\\ \frac{-4}{c^2}(c^2+a^2+b^2)y=\frac{-4}{c^2}(bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0)\\ y=\frac{bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0}{c^2+a^2+b^2}\\ \mathrm{Substitute}y=\frac{bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0}{c^2+a^2+b^2}\mathrm{in}\mathrm{equation}\left(4\right)\\ \left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}\left(\frac{bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0}{c^2+a^2+b^2}\right)=2x_0+\frac{2ad}{c^2}-\frac{2a{z}_0}{c^{}}\\ \\ \\ \end{array}$

$$\begin{gathered} \begin{array}{r}\frac{2}{c^2}\left\{abx+(c^2+b^2)\left(\frac{bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0}{c^2+a^2+b^2}\right)\right\}=\frac{2}{c^2}(c^2{y}_0+bd-bc{z}_0)\\ abx=c^2{y}_0+bd-bc{z}_0-(c^2+b^2)\left(\frac{bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0}{c^2+a^2+b^2}\right)\\ =\frac{c^2{a}^2{y}_0+c^2(c^2+b^2)y_0+a^{2}bd+(c^2+b^2)bd-a^{2}bc{z}_0-(c^2+b^2)bc{z}_0}{c^2+a^2+b^2}\end{array} \\ Continutionoftheabove. \\ =\frac{c^2{a}^2{y}_0+a^{2}bd-a^{2}bc{z}_0+(c^2+b^2)bc{x}_0-+a^2(c^2+b^2)y_0}{c^2+a^2+b^2} \\ =\frac{a^{2}bd-a^{2}bc{z}_0+c^{2}ab{x}_0+b^{2}ab{x}_0-a^2{b}^2{y}_0}{c^2+a^2+b^2} \\ =ab\left(\frac{ad-ac{z}_0+c^2{x}_0+b^2{x}_0-ab{y}_0}{c^2+a^2+b^2}\right) \\ x=\frac{ad-ac{z}_0+c^2{x}_0+b^2{x}_0-ab{y}_0}{c^2+a^2+b^2} \end{gathered}$$

$$\begin{gathered} Hence,theminimumormaximumexistat \\ \left(\frac{ad-ac{z}_0+c^2{x}_0+b^2{x}_0-ab{y}_0}{c^2+a^2+b^2},\frac{bd-ab{x}_0+c^2{y}_0+a^2{y}_0-bc{z}_0}{c^2+a^2+b^2}\right) \\ Todeterminethenaturepointcalculate{D}_{xx}(x,y),D_{yy}(x,y)and{D}_{xy}(x,y). \\ Differentiate(2)partiallywithrespecttox, \\ \frac{∂}{dx}{D}_x(x,y)=\frac{∂}{dx}\left\{2(x-x_0)-2.\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)\right\} \\ {D}_{xx}(x,y)=2\frac{∂}{∂x}(x-x_0)-2.\frac{a}{c}\frac{∂}{dx}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2+2\frac{a^2}{c^2}. \\ Differentiate\left(3\right)partiallywithrespecttoy, \\ \frac{∂}{dy}{D}_y(x,y)=\frac{∂}{dy}\left\{2(y-y_0)-2.\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)\right\} \\ {D}_{yy}(x,y)=2\frac{∂}{dy}(y-y_0)-2.\frac{b}{c}\frac{∂}{dx}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2+2\frac{b^2}{c^2}. \end{gathered}$$