3D Geometry 3 Question 48
####48. A variable plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ at a unit distance from origin cuts the coordinate axes at $A, B$ and $C$. Centroid $(x, y, z)$ satisfies the equation $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=K$. The value of $K$ is
(2005, 2M)
(a) 9
(b) 3
(c) $1 / 9$
(d) $1 / 3$
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Answer:
Correct Answer: 48. (a)
Solution:
- Since, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ cuts the coordinate axes at $A(a, 0,0), B(0, b, 0), C(0,0, c)$.
And its distance from origin $=1$
where, $P$ is centroid of triangle.
$ \begin{aligned} & \therefore & P(x, y, z) & =\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \\ & & x & =\frac{a}{3}, y=\frac{b}{3}, z=\frac{c}{3} \end{aligned} $
From Eqs. (i) and (ii),
$ \begin{aligned} & \frac{1}{9 x^{2}}+\frac{1}{9 y^{2}}+\frac{1}{9 z^{2}} & =1 \\ \text { or } & \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=9 & =K \\ \therefore K & =9 \end{aligned} $
