Three Dimensional Geometry Question 248
Question: Two systems of rectangular axes have the same origin. If a plane cuts them at distance a, b, c and a’, b’, c’ from the origin, then
[AIEEE 2003]
Options:
A) $ \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a{{’}^{2}}}+\frac{1}{b{{’}^{2}}}+\frac{1}{c{{’}^{2}}}=0 $
B) $ \frac{1}{a^{2}}+\frac{1}{b^{2}}-\frac{1}{c^{2}}+\frac{1}{a{{’}^{2}}}+\frac{1}{b{{’}^{2}}}-\frac{1}{c{{’}^{2}}}=0 $
C) $ \frac{1}{a^{2}}-\frac{1}{b^{2}}-\frac{1}{c^{2}}+\frac{1}{a{{’}^{2}}}-\frac{1}{b{{’}^{2}}}-\frac{1}{c{{’}^{2}}}=0 $
D) $ \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}-\frac{1}{a{{’}^{2}}}-\frac{1}{b{{’}^{2}}}-\frac{1}{c{{’}^{2}}}=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
Equation of planes be $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $ and $ \frac{x}{{{a}’}}+\frac{y}{{{b}’}}+\frac{z}{{{c}’}}=1 $ (Perpendicular distance on plane from origin is same)
\ $ | \frac{-1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}} |=| \frac{-1}{\sqrt{\frac{1}{{{{{a}’}}^{2}}}+\frac{1}{{{{{b}’}}^{2}}}+\frac{1}{{{{{c}’}}^{2}}}}} | $
\ $ \sum \frac{1}{a^{2}}-\sum \frac{1}{{{{{a}’}}^{2}}}=0 $ .
BETA
