Question:
A sphere of constant radius r through the origin intersects the coordinate axes in A, B and C. What is the locus of the centroid of the \(\triangle ABC\)?
मूल बिंदु से गुजरने वाला अचर त्रिज्या r का एक गोला निर्देशांक अक्षों को A, B और C पर प्रतिच्छेद करता है। \(\triangle ABC\) के केंद्रक (centroid) का बिंदु-पथ (locus) क्या है?
Options:
- \(9(x^2 + y^2 + z^2) = 4r^2\)
- \(x^2 + y^2 + z^2 = r^2\)
- \(x^2 + y^2 + z^2 = 4r^2\)
- \(3(x^2 + y^2 + z^2) = 2r^2\)
- \(9(x^2 + y^2 + z^2) = 4r^2\)
- \(x^2 + y^2 + z^2 = r^2\)
- \(x^2 + y^2 + z^2 = 4r^2\)
- \(3(x^2 + y^2 + z^2) = 2r^2\)
Correct Answer: a)
Explanation:
The sphere passes through origin and intercepts axes at \(A(a, 0, 0), B(0, b, 0), C(0, 0, c)\). Its diameter satisfies \(a^2 + b^2 + c^2 = (2r)^2 = 4r^2\).
Centroid \((x, y, z) = (a/3, b/3, c/3) \implies a = 3x, b = 3y, c = 3z\).
Substituting: \((3x)^2 + (3y)^2 + (3z)^2 = 4r^2 \implies 9(x^2 + y^2 + z^2) = 4r^2\).
गोले के व्यास के लिए \(a^2 + b^2 + c^2 = 4r^2\)। केंद्रक \((x, y, z) = (a/3, b/3, c/3)\)।
मान रखने पर: \(9(x^2 + y^2 + z^2) = 4r^2\)।