Question:
The distance between the point (1,2,3) and the sphere $x^{2} + y^{2} + z^{2} = 9$ is}
The distance between the point (1,2,3) and the sphere $x^{2} + y^{2} + z^{2} = 9$ is}
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Shortest distance = (Distance to center) - Radius. If the result is negative, the point is inside.
Updated On: Jul 3, 2026
- 0
- $3 + \sqrt{14}$
- $\sqrt{14} - 3$
- $3 - \sqrt{14}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The shortest distance from a point $P$ to a sphere is $|CP - R|$, where $C$ is the center and $R$ is the radius.
Step 2: Meaning
For $x^2 + y^2 + z^2 = 9$, the center $C = (0,0,0)$ and radius $R = 3$.
Step 3: Analysis
Distance $CP$ from $(0,0,0)$ to $(1,2,3)$ is $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
Step 4: Conclusion
Since $\sqrt{14} > 3$, the point is outside the sphere. The distance to the surface is $\sqrt{14} - 3$. Final Answer: (C)
The shortest distance from a point $P$ to a sphere is $|CP - R|$, where $C$ is the center and $R$ is the radius.
Step 2: Meaning
For $x^2 + y^2 + z^2 = 9$, the center $C = (0,0,0)$ and radius $R = 3$.
Step 3: Analysis
Distance $CP$ from $(0,0,0)$ to $(1,2,3)$ is $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
Step 4: Conclusion
Since $\sqrt{14} > 3$, the point is outside the sphere. The distance to the surface is $\sqrt{14} - 3$. Final Answer: (C)
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