Question

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

• A. 9
• B. 3
• C. 44570
• D. 44564

Answer 1

Answer: (A)

9

Answer 2

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
$\therefore$ \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