91影视

The given point is the origin and the constraint is$x+2y-3z=7.$

It is given that the point is $\left(0,0,0\right)$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$g(x,y,z)=x+2y鈭�3z鈭�7.$

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

$$\begin{gathered} \mathrm{鈭�}D(x,y,z)=2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k} \\ \mathrm{鈭�}\mathrm{g}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathbf{i}+2\mathbf{j}鈭�3\mathbf{k} \end{gathered}$$

Now, the system of the equation is

$$\begin{gathered} \mathrm{鈭�}D(x,y,z)=位\mathrm{鈭�}g(x,y,z) \\ 2x\mathit{i}+2y\mathit{j}+2z\mathit{k}=位(\mathit{i}+2\mathit{j}鈭�3\mathit{k}) \end{gathered}$$

Here,$2x=位,2y=2位\text{and}2z=鈭�3位.$

The values of $位,$we get are$2x=\frac{2y}{2}=\frac{2z}{鈭�3}=位.$

So, $y=2x\text{and}z=鈭�3x.$

Substitute the above values in the given constraint,

role="math" localid="1649939263170" $$\begin{gathered} x+2y-3z=7 \\ x+2\left(2x\right)-3\left(-3x\right)=7 \\ x+4x+9x=7 \\ 14x=7 \\ x=\frac{1}{2} \\ \mathrm{So},y=1,z=鈭�\frac{3}{2} \end{gathered}$$

Thus, the points where the extreme value may appear are$\left(\frac{1}{2},1,鈭�\frac{3}{2}\right).$

To find the point closest to the specified point, let's put the points in the function $D(x,y,z)=x^2+y^2+z^2.$

So, when the points are $\left(\frac{1}{2},1,鈭�\frac{3}{2}\right)$the value is $\frac{7}{2}.$

Thus, the points on the given surface closest to$\left(0,0,0\right)a\mathrm{re}\left(\frac{1}{2},1,鈭�\frac{3}{2}\right).$