Q. 16 - Calculus

Question

Let $\mathcal{P}$ be the plane ax + by + c z = d, N = (a, b ,c)  be the normal vector to $\mathcal{P}$, R be a point on $\mathcal{P}$, and P be the point $(x_0, y_0,z_0).$ Show that the distance formula we derived for computing the distance from point P to plane $\mathcal{P}$ in Chapter 10, $\frac{\left|N.\overrightarrow{RP}\right|}{||N||}$, is equivalent to the distance formula we derived in Example 4. That is, show that $\frac{\left|N.\overrightarrow{RP}\right|}{||N||}=\frac{\left|a{x}_0+b{y}_0+c{z}_0-d\right|}{\sqrt{a^2+b^2+c^2}}.$

Step-by-Step Solution

Answer

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Step 1. Given

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

Step 2. Calculation

Step 3. Calculation

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{function} , \\ D( x ,y ) = {( x - x_0)}^2+ {( y- y_0)}^2+ {(\frac{d- ax -by }{c}-z_0)}^2....\left(1\right) \\ \mathrm{First} \mathrm{show} \mathrm{that} \mathrm{the} \mathrm{point} , \\ \left(\frac{ad- b{y}_0-ac{z}_0+ b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd- ab{x}_0-bc{z}_0+ a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2} \right) \\ \mathrm{gives} \mathrm{absolute} \mathrm{minimum} \mathrm{for} \mathrm{the} \mathrm{given} \mathrm{function} . \\ \mathrm{Differentiate} \left(1\right)\hspace{0.17em}\mathrm{partially} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} , \\ \frac{\partial }{\partial x}D( x ,y ) = \frac{\partial }{\partial x}\{ {( x - x_0)}^2+ {( y- y_0)}^2+ {(\frac{d- ax -by }{c}-z_0)}^2\} \\ {D}_x( x , y) =\frac{\partial }{\partial x}{( x - x_0)}^2+\frac{\partial }{\partial x}{( y- y_0)}^2+\frac{\partial }{\partial x} {(\frac{d- ax -by }{c}-z_0)}^2. \\ = 2( x - x_0) +2 (\frac{d- ax -by }{c}-z_0)\frac{\partial }{\partial x} (\frac{d- ax -by }{c}-z_0). \\ = 2( x - x_0) +2 (\frac{d- ax -by }{c}-z_0)\{\frac{\partial }{\partial x} \left(\frac{d- ax -by }{c}\right)-\frac{\partial }{\partial x}{z}_0\} \\ = 2( x - x_0) +2 (\frac{d- ax -by }{c}-z_0)(\frac{-a}{c}-0) = 2( x - x_0) -2\frac{a}{c} (\frac{d- ax -by }{c}-z_0)......\left(2\right) \end{gathered}$$

Step 4. Calculation

$$\begin{gathered} \mathrm{Differentiate} \left(1\right)\hspace{0.17em}\mathrm{partially} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{y}, \\ \frac{\partial }{\partial y}D( x ,y ) = \frac{\partial }{\partial y}\{ {( x - x_0)}^2+ {( y- y_0)}^2+ {(\frac{d- ax -by }{c}-z_0)}^2\} \\ {D}_y( x , y) =\frac{\partial }{\partial y}{( x - x_0)}^2+\frac{\partial }{\partial y}{( y- y_0)}^2+\frac{\partial }{\partial y} {(\frac{d- ax -by }{c}-z_0)}^2. \\ = 2( y- y_0) +2 (\frac{d- ax -by }{c}-z_0)\frac{\partial }{\partial y} (\frac{d- ax -by }{c}-z_0). \\ = 2( y- y_0) +2 (\frac{d- ax -by }{c}-z_0)\{\frac{\partial }{\partial y} \left(\frac{d- ax -by }{c}\right)-\frac{\partial }{\partial y}{z}_0\} \\ = 2( y- y_0) +2 (\frac{d- ax -by }{c}-z_0)(\frac{-b}{c}-0) = 2( x - x_0) -2\frac{b}{c} (\frac{d- ax -by }{c}-z_0).....\left(3\right) \end{gathered}$$

Step 5. Calculating stationary point

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

Step 6. Calculation

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

Step 7.

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

Step 8.

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

Step 9.

$$\begin{gathered} Hence , the minimum or maximum exist at \\ \left(\frac{ad-ac{z}_0+c^2{x}_0+b^2{x}_0-ab{y}_0}{c^2+a^2+b^2},\frac{bd-ab{x}_0+c^2{y}_0+a^2{y}_0-bc{z}_0}{c^2+a^2+b^2}\right) \\ To determine the nature point calculate D_{xx}(x , y) ,D_{yy}(x , y) and D_{xy}(x , y) . \\ Differentiate (2 ) partially with respect to x , \\ \frac{\partial }{dx}{D}_x(x , y)=\frac{\partial }{dx}\left\{2(x - x_0)-2.\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)\right\} \\ D_{xx}(x , y) =2\frac{\partial }{\partial x}(x - x_0)-2.\frac{a}{c}\frac{\partial }{dx}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2 +2\frac{a^2}{c^2}. \\ Differentiate \left(3\right) partially with respect to y , \\ \frac{\partial }{dy}{D}_y(x , y)=\frac{\partial }{dy}\left\{2(y - y_0)-2.\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)\right\} \\ D_{yy}(x , y) =2\frac{\partial }{dy}(y - y_0)-2.\frac{b}{c}\frac{\partial }{dx}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2 +2\frac{b^2}{c^2}. \end{gathered}$$