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. Band 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
- 9
- 3
- 44570
- 44564
The Correct Option is A
Solution and Explanation
A (a, 0,0), B (0, b,0), C (0,0, c).
And its distance from origin = 1
$\therefore$ $\hspace15mm \frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=1$
or $\hspace15mm\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1 \hspace35mm ...(i) $
where, P is centroid of triangle.
$\therefore$ $\hspace10mm P(x,y,z) = \bigg(\frac{a\, +\, 0\, +\, 0}{3},\frac{0\, +\, b\, +\, 0}{3},\frac{0\, +\, 0\, +\, c}{3}\bigg)$
$\Rightarrow $ $\hspace23mm x =\frac{a}{3},y =\frac{b}{3},z =\frac{c}{3} \hspace20mm ...(ii) $
From Eqs. (i) and (ii),
$\hspace30mm$ $ \frac{1}{9x^2}+ \frac{1}{9y^2}+ \frac{1}{9z^2} = 1 $
or $\hspace25mm$ $ \frac{1}{x^2}+ \frac{1}{y^2}+ \frac{1}{z^2} = 9 =K$
$\therefore$ $\hspace50mm K = 9 $
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