A variable plane is at a distance of 6 units | vedclass

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(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Given that the distance from the origin is $6$ units. Using the fo

Vedclass · Accepted Answer

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{4}$

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(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Given that the distance from the origin is $6$ units. Using the formula for the distance of a plane from the origin,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 6$. Squaring both sides,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{36} \quad (i)$. The coordinates of $A, B$,and $C$ are $(a, 0, 0), (0, b, 0)$,and $(0, 0, c)$ respectively. The centroid $(x, y, z)$ of $	riangle ABC$ is given by $x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}$. Thus,$a = 3x, b = 3y, c = 3z$. Substituting these values into equation $(i)$,we get $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{36}$. $\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{36}$. Multiplying by $9$,we get $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{36} = \frac{1}{4}$.

$A$ variable plane is at a distance of $6$ units from the origin. If it meets the coordinate axes in $A, B$,and $C$,then the equation of the locus of the centroid of the $\triangle ABC$ is

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