Three Dimensional Geometry Question 312
Question: A variable plane which remains at a constant distance 3p from the origin cut the coordinate axes at A, B and C. The locus of the centroid of triangle ABC is
Options:
A) $ {x^{-1}}+{y^{-1}}+{z^{-1}}={p^{-1}} $
B) $ {x^{-2}}+{y^{-2}}+{z^{-2}}={p^{-2}} $
C) $ x+y+z=p $
D) $ x^{2}+y^{2}+z^{2}=p^{2} $
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Answer:
Correct Answer: B
Solution:
[b] Let equation of the variable plane be $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $ This meets the coordinate axes at A (a, 0, 0), B (0, b, 0) and C (0, 0, c). Let $ P(\alpha ,\beta ,\gamma ) $ be the centroid of the $ \Delta ABC. $ Then $ \alpha =\frac{a+0+0}{3},\beta =\frac{0+b+0}{3},\gamma =\frac{0+0+c}{3} $
$ \therefore a=3\alpha ,b=3\beta ,c=3\gamma $ (2) Plane (1) is at constant distance 3p form the origin, so $ 3p=\frac{| \frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1 |}{\sqrt{{{( \frac{1}{a} )}^{2}}+{{( \frac{1}{b} )}^{2}}+{{( \frac{1}{c} )}^{2}}}} $
$ \Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{9p^{2}} $ (3) Form (2) and (3), we get $ \frac{1}{9{{\alpha }^{2}}}+\frac{1}{9{{\beta }^{2}}}+\frac{1}{9{{\gamma }^{2}}}=\frac{1}{9p^{2}} $
$ \Rightarrow {{\alpha }^{-2}}+{{\beta }^{-2}}+{{\gamma }^{-2}}={p^{-2}} $ Generalizing $ \alpha ,\beta ,\gamma , $ locus of centroid P $ P(\alpha ,\beta ,\gamma ) $ is $ {x^{-2}}+{y^{-2}}+{z^{-2}}={p^{-2}} $
BETA
