We consider two spheres. The sphere of radius 3 centered at the origin (0, 0, 0). The second sphere is tangent to the first sphere and is centered at A(3, 5, 2). Find the point of intersection P of...

We consider two spheres. The sphere of radius 3 centered at the origin (0, 0, 0).

The second sphere is tangent to the first sphere and is centered at A(3, 5, 2). Find the point of intersection P of these spheres.

Hint: The points A, P and the origin are on the same line.

Asked on by swimmers

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

Using the hint, the point of intersection `P` is 3 units away from the origin `O` and in the same direction as the vector `vec(OA)=(3,5,2).` The length of `vec(OA)` is given by

`|vec(OA)|=sqrt(3^2+5^2+2^2)=sqrt(38),` and the vector parallel to this with length 3 is just

`3/sqrt(38)vec(OA)=(9/sqrt(38),15/sqrt(38),6/sqrt(38)),`

so `(9/sqrt(38),15/sqrt(38),6/sqrt(38)).`

 

Another (related) way to solve it is to realize that `P` must lie on the line described parametrically by `(3t,5t,2t).` We need to find the value of `t` such that `P` is 3 units away from the origin. In other words, we need to solve

`3=sqrt((3t)^2+(5t)^2+(2t)^2)=sqrt(38t^2),` and we get `t=3/sqrt(38).` Plugging this in gives, as before,

`(9/sqrt(38),15/sqrt(38),6/sqrt(38)).`

 

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

The last line in both solutions should read

`P=(9/sqrt(38),15/sqrt(38),6/sqrt(38)).`  ``

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