Find the point on the plane ax + by + cz = d at minimum distance from the origin using the method of lagrange multipliers.

The distance between some point [tex](x,y,z)[/tex] and the origin is given by

[tex]f(x,y,z)=\sqrt{x^2+y^2+z^2}[/tex]

so this is the function we're trying to minimize. But notice that [tex]f(x,y,z)[/tex] and [tex]f(x,y,z)^2[/tex] attain their critical points at the same [tex](x,y,z)[/tex], so we can solve the same problem by minimizing [tex]x^2+y^2+z^2[/tex] instead.

So let's take the Lagrangian to be

[tex]L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(ax+by+cz-d)[/tex]

with partial derivatives (set equal to 0)

[tex]L_x=2x+a\lambda=0[/tex]
[tex]L_y=2y+b\lambda=0[/tex]
[tex]L_z=2z+c\lambda=0[/tex]
[tex]L_\lambda=ax+by+cz-d=0[/tex]

Now, notice that

[tex]aL_x+bL_y+cL_z=2(ax+by+cz)+(a^2+b^c+c^2)\lambda=0[/tex]
[tex]\implies\lambda=-\dfrac{2d}{a^2+b^2+c^2}[/tex]

and we can use this to solve for [tex]x,y,z[/tex]. We get

[tex]x=\dfrac{ad}{a^2+b^2+c^2}[/tex]
[tex]y=\dfrac{bd}{a^2+b^2+c^2}[/tex]
[tex]z=\dfrac{cd}{a^2+b^2+c^2}[/tex]

At this critical point, we get a minimum distance of

[tex]\sqrt{\left(\dfrac{ad}{a^2+b^2+c^2}\right)^2+\left(\dfrac{bd}{a^2+b^2+c^2}\right)^2+\left(\dfrac{cd}{a^2+b^2+c^2}\right)^2}=\sqrt{\dfrac{d^2}{a^2+b^2+c^2}}[/tex]