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This is a silly question but I'm stuck.. I'm given the sphere $(C):(x-1)^2+(y-2)^2+(z-3)^2=1$ with radius $r=1$ and center $K(1,2,3)$ and the plane $(P):x+y+z=6+\sqrt3$

I proved that they are tangent, I need to find their contact point

Any hints?

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  • $\begingroup$ you could use the general point $X$ on plane $P$ and then calculate the distance $\overline{XK}$. The point of contact then would be the minimum of that function. $\endgroup$ – Dr. Richard Klitzing May 12 '18 at 11:27
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Recall that the radius of the sphere is perpendicular to the plane at their common point. So you need only find the projection of K on the given plane. To do that, find the vector that is perpendicular to P and then find a straight line parallel to it that passes from K. If you need more help, edit your progress first.

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  • $\begingroup$ Ok I thing I got it! $u=(1,1,1)$ is perpendicular to $P$ so the line that passes through $K$ perpendicular to $P$ is $x=t+1,y=t+2,z=t+3$ $(P): t+1+t+2+t+3=6+\sqrt3$ so $t=\sqrt3/3$ So $x=\sqrt3/3+1, y=\sqrt3/3+2, z=\sqrt3/3+3$ And the contact point is $(\sqrt3/3+1,\sqrt3/3+2,\sqrt3/3+3)$ Is this correct? $\endgroup$ – VakiPitsi May 12 '18 at 11:37
  • $\begingroup$ Yes, very well! $\endgroup$ – George K May 12 '18 at 11:45
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The contact point will be the closest point on the plane from the center of the sphere. So if the point is $(x,y,z)$ then we have to minimize the distance $(x-1)^2+(y-2)^2+(z-3)^2$ subject to the condition that the point belongs to the plane. Using Lagrange multiplier it is done quickly.

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