91Ó°ÊÓ

The given point is the origin and the constraint is$ax+by+cz=d.$

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$g(x,y,z)=ax+by+cz−d.$

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

$$\begin{gathered} \mathrm{∇}D(x,y,z)=2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k} \\ \mathrm{∇}\mathrm{g}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathrm{a}\mathbf{i}+\mathrm{b}\mathbf{j}+\mathrm{c}\mathbf{k} \end{gathered}$$

Now, the system of the equation is

$$\begin{gathered} \mathrm{∇}D(x,y,z)=\mathrm{λ}\mathrm{∇}g(x,y,z) \\ 2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}=\mathrm{λ}(a\mathbf{i}+b\mathbf{j}+c\mathbf{k}) \end{gathered}$$

Here,$2x=a\mathrm{λ},2y=b\mathrm{λ}\text{and}2z=c\mathrm{λ}.$

The values of $\mathrm{λ},$we get are$\frac{2x}{a}=\frac{2y}{b}=\frac{2z}{c}=\mathrm{λ}.$

So, $y=\frac{b}{a}x\text{and}z=\frac{c}{a}x.$

Substitute the above values in the given constraint,

$$\begin{gathered} ax+by+cz=d \\ ax+b\left(\frac{b}{a}x\right)+c\left(\frac{c}{a}x\right)=d \\ x(a^2+b^2+c^2)=ad \\ x=\frac{ad}{a^2+b^2+c^2} \\ \mathrm{So},y=\frac{bd}{a^2+b^2+c^2},z=\frac{cd}{a^2+b^2+c^2} \end{gathered}$$

Thus, the points where the extreme value may appear are$(\frac{ad}{a^2+b^2+c^2},\frac{bd}{a^2+b^2+c^2},\frac{cd}{a^2+b^2+c^2}).$

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 $\left(\frac{ad}{a^2+b^2+c^2},\frac{bd}{a^2+b^2+c^2},\frac{cd}{a^2+b^2+c^2}\right)$the value is $\frac{d^2}{a^2+b^2+c^2}.$

Thus, the points on the given surface closest to$\left(0,0,0\right)\mathrm{are}\left(\frac{ad}{a^2+b^2+c^2},\frac{bd}{a^2+b^2+c^2},\frac{cd}{a^2+b^2+c^2}\right).$