Three Dimensional Geometry Question 57
Two systems of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a’, b’, c’ respectively from the origin, then $ \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=k( \frac{1}{a{{’}^{2}}}+\frac{1}{b{{’}^{2}}}+\frac{1}{c{{’}^{2}}} ) $ , where k is equal to
Options:
1
2
4
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let a, b, c, be the intercepts when (Ox, Oy, Oz) are taken as axes, then the equation of the plane is $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $ Let a, b, c be the intercepts when (OX, OY, OZ) are taken as axes; then in this case equation of the same plane is $ \frac{X}{a}+\frac{Y}{b}+\frac{Z}{c}=1 $ (i) Now, eqs. (i) and (ii) are equation of the same plane and in both the cases the origin is same. Hence, length of the perpendicular drawn from the origin to the plane in both the case must be the same
$ \therefore \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}}}}} $
$ \Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{a{{’}^{2}}}+\frac{1}{b{{’}^{2}}}+\frac{1}{c{{’}^{2}}} $
$ \therefore k=1 $
BETA
