A variable plane is at a constant distance h | vedclass

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(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. The points where it meets the coordinate axes are $A(a, 0, 0)$, $B

Vedclass · Accepted Answer

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{h^2}$

Vedclass · Answer

(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. The points where it meets the coordinate axes are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$.The distance of this plane from the origin $(0, 0, 0)$ is given as $h$. The formula for the distance is $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = h$, which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{h^2}$.Let $(x, y, z)$ be the coordinates of the centroid of $	riangle ABC$. Then $x = \frac{a}{3}$, $y = \frac{b}{3}$, and $z = \frac{c}{3}$.This gives $a = 3x$, $b = 3y$, and $c = 3z$.Substituting these into the distance equation:$\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{h^2}$$\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{h^2}$$\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{h^2}$

$A$ variable plane is at a constant distance $h$ from the origin and meets the coordinate axes in $A, B, C$. The locus of the centroid of $\triangle ABC$ is

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