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

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Equation of circle:

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

3. 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
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 Recognitions: Homework Help $$\sqrt{(x-2)^2+(y-2)^2+(z-2)^2}=1.5$$ 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?

 Quote by rock.freak667 $$\sqrt{(x-2)^2+(y-2)^2+(z-2)^2}=1.5$$ 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.