91影视

The given cone is $z^2=x^2+y^2$and the plane is$x+2y=6.$

It is given that we have to find the point closest to the origin so, the distance of a point from the origin is $\sqrt{x^2+y^2+z^2}.$

Let the function be the square of the above distance $D(x,y,z)=x^2+y^2+z^2$and the constraint as $g(x,y,z)=x^2+y^2鈭�z^2\mathrm{and}h(x,y,z)=x+2y鈭�6.$

By using the Lagrange's method, the gradient of the above functions is,

$\begin{array}{rl}\mathrm{鈭�}D(x,y,z)=& 2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}\\ \mathrm{鈭�}g(x,y,z)=& 2x\mathbf{i}+2y\mathbf{j}-2\mathrm{z}\mathbf{k}\\ \mathrm{鈭�}h(x,y,z)=& \mathbf{i}+2\mathbf{j}\end{array}$

Now, the system of the equation is

$$\begin{gathered} \mathrm{鈭�}D(x,y,z)=位\mathrm{鈭�}g(x,y,z)+渭\mathrm{鈭�}h(x,y,z) \\ \begin{array}{rl}2x\mathrm{i}+2y\mathrm{j}+2z\mathrm{k}& =位(2x\mathrm{i}+2y\mathrm{j}鈭�2z\mathrm{k})+渭(\mathrm{i}+2\mathrm{j})\\ 2x\mathrm{i}+2y\mathrm{j}+2z\mathrm{k}& =(2位x+渭)\mathrm{i}+(2位y+2渭)\mathrm{j}鈭�2位z\mathrm{k}\end{array} \end{gathered}$$

Here,

$$\begin{gathered} 2x=2位x+渭.......\left(a\right) \\ \begin{array}{rl}2y& =2位y+2渭\\ y& =位y+渭.........\left(b\right)\\ 2z=& -2位z\\ z=& -位.........\left(c\right)\end{array} \end{gathered}$$

The values of role="math" localid="1649987568353" $渭,$we get are

role="math" localid="1649987643384" $$\begin{gathered} 渭=2x-2\left(-z\right)x \\ 渭=2x+2xz......\left(d\right) \\ 渭=y+yz.........\left(e\right) \end{gathered}$$

Now, equate equation (e) and (f)

$$\begin{gathered} 2x+2xz=y+yz \\ 2x+2xz-y-yz=0 \\ 2x(1+z)-y(1+z)=0 \\ \left(2x-y\right)\left(1+z\right)=0 \\ z=0\mathrm{and}y=2x \end{gathered}$$

Substitute the above values in the given constraints,

$$\begin{gathered} \mathrm{If}z=0 \\ {x}^2+y^2=0 \\ {x}^2=-y^2.....\left(g\right) \\ x=6-2y \\ \mathrm{Substitute}\mathrm{above}\mathrm{equation}\mathrm{in}\mathrm{equstion}\left(\mathrm{g}\right) \\ {\left(6-2\mathrm{y}\right)}^2=-{\mathrm{y}}^2 \\ 36+4{\mathrm{y}}^2-24\mathrm{y}=-{\mathrm{y}}^2 \\ 36+5{\mathrm{y}}^2-24\mathrm{y}=0 \end{gathered}$$

Through the above equation, we get complex roots.

Now, if $y=2x$

$$\begin{gathered} x^2+y^2=z^2 \\ {x}^2+{\left(2x\right)}^2=z^2 \\ 5x^2=z^2......\left(h\right) \\ x+2\left(2x\right)=6 \\ 5x=6 \\ x=\frac{6}{5} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{equation}\left(\mathrm{h}\right) \\ 5{\left(\frac{6}{5}\right)}^2={\mathrm{z}}^2 \\ \frac{6}{\sqrt{5}}=\mathrm{z} \\ \mathrm{And} \\ {x}^2+y^2=z^2 \\ {\left(\frac{6}{5}\right)}^2+y^2={\left(\frac{6}{\sqrt{5}}\right)}^2 \\ {y}^2=\frac{36}{5}-\frac{36}{25} \\ {y}^2=\frac{144}{25} \\ y=\frac{12}{5} \end{gathered}$$

Thus, the points in ${\mathrm{鈩�}}^3$closest to the origin is$\left(\frac{6}{5},\frac{12}{5},\frac{6}{\sqrt{5}}\right).$