How to verify a position vector points to a point inside a sphere?

[yungman]

Homework Statement

Given a sphere center at (2,2,2) with radius of 1.5, given a vector A=<3,2,1>. How do I verify A point to a point inside the sphere?

Homework Equations

Equation of circle:

\sqrt{(x-2)^2+(y-2)^2+(z-2)^2}=1.5

The Attempt at a Solution

I know A point to a point inside the sphere.

\sqrt{(3-2)^2+(2-2)^2+(1-2)^2}=\sqrt{2}\;\hbox { smaller than 1.5}

Is any point that give number smaller than 2.25 indicate the point is inside the sphere? can you explain why?

Thanks

Alan

 

[rock.freak667]

<br /> \sqrt{(x-2)^2+(y-2)^2+(z-2)^2}=1.5<br />

This equation will give any point on the surface of the sphere since the distance from the center to the surface is constant.

If your point is before the surface, would the distance be more or less than the radius?

 

[yungman]

<br /> \sqrt{(x-2)^2+(y-2)^2+(z-2)^2}=1.5<br />

This equation will give any point on the surface of the sphere since the distance from the center to the surface is constant.

If your point is before the surface, would the distance be more or less than the radius?

It would be more than the radius if the point is outside the sphere.

I have to think a little about this.

Thanks

 

[hunt_mat]

The way to go about this question is to examine the distance between the centre of the sphere and the point defined by the vector. If the distance is less than the radius of te sphere then you know that the point is inside the sphere.

 

[yungman]

Got it. thanks both of you for the help.

Alan