If a variable plane,at a distance of 3 units | vedclass

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(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $A = (a, 0, 0)$,$B = (0, b, 0)$,and $C = (0, 0, c)$.The dista

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $A = (a, 0, 0)$,$B = (0, b, 0)$,and $C = (0, 0, c)$.The distance of this plane from the origin $(0, 0, 0)$ is given as $3 \ units$. Thus,$\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 3$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9}$.The centroid $(x, y, z)$ of $\Delta ABC$ is given by $x = \frac{a+0+0}{3}$,$y = \frac{0+b+0}{3}$,and $z = \frac{0+0+c}{3}$.Therefore,$a = 3x$,$b = 3y$,and $c = 3z$.Substituting these values into the distance equation: $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{9}$.$\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{9}$.Multiplying by $9$,we get $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1$.

If a variable plane,at a distance of $3 \ units$ from the origin,intersects the coordinate axes at $A, B$,and $C$,then the locus of the centroid of $\Delta ABC$ is

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