x,y,z intercepts of plane tangent to a sphere

[SigepBrandon]

The problem reads:
Find the x,y and z intercepts of the plane tangent to the sphere of radius sqrt(14), with center at the origin, at the point (1,2,3).

well...
I believe f(x,y,z) = x[sup:26f4zqev]2[/sup:26f4zqev]+y[sup:26f4zqev]2[/sup:26f4zqev]+z[sup:26f4zqev]2[/sup:26f4zqev]-14 to be correct.

I also found the partials f[sub:26f4zqev]x[/sub:26f4zqev](x,y,z)=2x, f[sub:26f4zqev]y[/sub:26f4zqev](x,y,z)=2y, and f[sub:26f4zqev]z[/sub:26f4zqev](x,y,z)=2z, evaluated the partials at the point (1,2,3) which gave me f[sub:26f4zqev]x[/sub:26f4zqev](1,2,3)=2, f[sub:26f4zqev]y[/sub:26f4zqev](1,2,3)=4, f[sub:26f4zqev]z[/sub:26f4zqev](1,2,3)=6.

And I think the point (1,2,3) is also the normal vector to the tangent plane with magnitude sqrt(14).

I'm not really sure on what to do next, my guess is that the equation for the line is

x+2y[sup:26f4zqev]2[/sup:26f4zqev]+3z[sup:26f4zqev]2[/sup:26f4zqev]=14 and then I solve for the intercepts?

so x-int= sqrt(14), y-int = sqrt(7) z-int= sqrt(14/3)?

Any guidance would be much appreciated.

 

[galactus]

You have the partials.

\(\displaystyle f_{x}=2x, \;\ f_{y}=2y, \;\ f_{z}=2z\)

At (1,2,3) the partials are then:

\(\displaystyle f_{x}(1,2,3)=2(1)=2\)

\(\displaystyle f_{y}(1,2,3)=2(2)=4\)

\(\displaystyle f_{z}(1,2,3)=2(3)=6\)

The normal is \(\displaystyle n=2i+4j+6k\)

So, the equation of the tangent plane is \(\displaystyle 2(x-1)+4(y-2)+6(z-3)=0\)

\(\displaystyle 2x+4y+6z-28=0\Rightarrow x+2y+3z=14\)

The x intercept can be found when y and z are 0, and so on.

It would appear your intercepts are correct except for the sqrt.

The equation of a line will not have squared terms.

 

[SigepBrandon]

thanks! I see my errors and better understand the solution. much appreciated.