Three Dimensional Geometry Question 271
Question: A variable plane at a constant distance p from origin meets the co-ordinates axes in $ A,B,C $ . Through these points planes are drawn parallel to co-ordinate planes. Then locus of the point of intersection is
Options:
A) $ \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\frac{1}{p^{2}} $
B) $ x^{2}+y^{2}+z^{2}=p^{2} $
C) $ x+y+z=p $
D) $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=p $
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of plane is, $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $
$ {a,b,c $ respectively are intercepts on $ x,y,z $ axes}
Then $ \frac{abc}{\sqrt{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}}=p $
$ ,\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}} $
Therefore locus of the point (x, y, z) is
$ \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\frac{1}{p^{2}} $ .
BETA
